Question 13 Marks
If the speed of light is $3.0 \times 10^8 \mathrm{~ms}^{-1}$, calculate the distance covered by light in 2.00 ns .
AnswerAccording to the question:
Time taken to cover the distance $=2.00 \mathrm{~ns}$
$=2.00 \times 10^{-9} \mathrm{~s}$
Speed of light $=3.0 \times 10^8 \mathrm{~ms}^{-1}$
Distance travelled by light in 2.00 ns
$=\text { Speed of light } \times \text { Time taken }$
$=\left(3.0 \times 10^8 \mathrm{~ms}^{-1}\right)\left(2.00 \times 10^{-9} \mathrm{~s}\right)$
$=6.00 \times 10^{-1} \mathrm{~m}$
$=0.600 \mathrm{~m}$
View full question & answer→Question 23 Marks
The following data are obtained when dinitrogen and dioxygen react together to form different compounds:
Fill in the blanks in the following conversions:
- 1km = ...................... mm = ...................... pm.
- 1mg = ...................... kg = ...................... ng.
- 1mL = ...................... L = ...................... $\text{dm}^3$.
Answer
- $1\text{km}= 1\text{km}\times\frac{1000\text{m}}{1\text{km}}\times\frac{100\text{cm}}{1\text{m}}\times\frac{10\text{mm}}{1\text{cm}}$
$\therefore1\text{km}=10^6\text{mm}$
$1\text{km}= 1\text{km}\times\frac{1000\text{m}}{1\text{km}}\times\frac{1\text{pm}}{10^{-12}\text{m}}$
$\therefore1\text{km}=10^{15}\text{pm}$
Hence, $1\text{km} = 10^6\text{mm} = 10^{15}\text{pm}$
- $1\text{mg} = 1\text{mg}×\frac{1\text{g}}{1000\text{mg}}\times\frac{1\text{kg}}{1000\text{g}}$
$⇒ 1 \text{mg} = 10^{–6} \text{kg}$
$1\text{mg} = 1\text{mg}×\frac{1\text{g}}{1000\text{mg}}\times\frac{1\text{ng}}{10^{-9}\text{g}}$
$⇒ 1 \text{mg} = 10^{–6} \text{ng}$
$\therefore1\text{mg}=10^{-6}\text{kg}=10^{6}\text{ng}$
- $1 \text{mL} = 1 \text{mL} × \frac{1\text{L}}{1000\text{mL}}$
$⇒ 1 \text{mL} = 10^{–3}\text{L}$
$1 \text{mL} = 1 \text{cm}^3 = 1 \text{cm}^3\frac{1\text{dm}\times1\text{dm}\times1\text{dm}}{10\text{cm}\times10\text{cm}\times10\text{cm}}$
$⇒ 1 \text{mL} = 10^{–3}\text{dm}^3$
$\therefore 1 \text{mL} = 10^{–3} \text{L} = 10^{–3} \text{dm}^3$ View full question & answer→Question 33 Marks
How are $0.50 \mathrm{~mol} \mathrm{Na}_2 \mathrm{CO}_3$ and $0.50 \mathrm{M} \mathrm{Na}_2 \mathrm{CO}_3$ different?
AnswerMolar mass of $\mathrm{Na}_2 \mathrm{CO}_3=(2 \times 23)+12.00+(3 \times 16)$
$=106 \mathrm{~g} \mathrm{~mol}^{-1}$
Now, 1 mole of $\mathrm{Na}_2 \mathrm{CO}_3$ means 106 g of $\mathrm{Na}_2 \mathrm{CO}_3$.
$\therefore 0.5 \mathrm{~mol}$ of $\mathrm{Na}_2 \mathrm{CO}_3=\frac{106 \mathrm{~g}}{1 \mathrm{~mole}} \times 0.5 \mathrm{~mol} \mathrm{Na}_2 \mathrm{CO}_3$
$=53 \mathrm{~g} \mathrm{Na}_2 \mathrm{CO}_3$
$\Rightarrow 0.50 \mathrm{M}$ of $\mathrm{Na}_2 \mathrm{CO}_3=0.50 \mathrm{~mol} / \mathrm{L} \mathrm{Na}_2 \mathrm{CO}_3$
Hence, 0.50 mol of $\mathrm{Na}_2 \mathrm{CO}_3$ is present in 1 L of water or 53 g of $\mathrm{Na}_2 \mathrm{CO}_3$ is present in 1 L of water.
View full question & answer→Question 43 Marks
Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
$N_{2(g)} + H_{2(g)} → 2NH_{3(g)}$
Calculate the mass of ammonia produced if $2.00 \times 10^3g$ dinitrogen reacts with $1.00 \times 10^3g$ of dihydrogen.
AnswerBalancing the given chemical equation,
$N_{2(g)}+H_{2(g)} \rightarrow 2 NH_{3(g)}$
From the equation, 1 mole ( 28 g ) of dinitrogen reacts with 3 mole $(6 g)$ of dihydrogen to give $2 mole(34 g)$ of ammonia.
$\Rightarrow 2.00 \times 10^3 g$ of dinitrogen will react with $\frac{6 g}{28 g} \times 2.00 \times 10^3 g$ dihydrogen i.e.,
$2.00 \times 10^3 g$ of dinitrogen will react with 428.6 g of dihydrogen.
Given,
Amount of dihydrogen $=1.00 \times 10^3 g$
Hence, $N _2$ is the limiting reagent.
$\therefore 28 g$ of $N _2$ produces 34 g of $NH _3$.
Hence, mass of ammonia produced by 2000 g of $N _2=\frac{34 g}{28 g} \times 2000 g$
$=2428.57 g .$
View full question & answer→Question 53 Marks
Calculate the mass of sodium acetate $\left(\mathrm{CH}_3 \mathrm{COONa}\right)$ required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is $82.0245 \mathrm{~g} \mathrm{~mol}^{-1}$.
Answer0.375 M aqueous solution means that 1000ml of the solution contain sodium acetate = 0.375 mole.
$\therefore$ 500ml of the solution contain sodium acetate $=\frac{0.375}{2}$ mole
Molar mass of sodium acetate = 82.0245g mol-1
$\therefore$ Mass of sodium acetate required $=\frac{0.375}{2}$ mole, × 82.0245g mol-1 = -15.380g.
View full question & answer→Question 63 Marks
Calculate the volume of 02 at STP liberated by heating 12.25 g of $\mathrm{KClO}_3$ (At. wt. of $\mathrm{K}=39, \mathrm{Cl}=35.5,0=16 \mathrm{u}$ )
Answer$2 \mathrm{KClO}_3 \xrightarrow{\text { heat }} 2 \mathrm{KCl}+3 \mathrm{O}_2$
Since $2 \times 122.5 \mathrm{~g}$ of $\mathrm{KClO}_3$ given $3 \times 22.4 \mathrm{~L}$ of $\mathrm{O}_2$ at STP
Therefore, 12.25 g of $\mathrm{KClO}_3$ given
$\frac{3 \times 22.4}{2 \times 122.5} \times 12.25$
$=3.36 \mathrm{~L}$ of $\mathrm{O}_2$ at STP.
View full question & answer→Question 73 Marks
The average molar mass of a mixture of methane ($CH_4$) and ethane ($C_2H_4$) present in the ratio of a : b is found to be 20.0g $mol^{-1}$. If the ratio were reversed, what would be the molar mass of the mixture?
AnswerMolar mass of $CH_4 = 16g mol^{-1}$
Molar mass of $C_2H_4 = 28g mol^{-1}$
When they are present in the a : b, their average molar mass
$=\frac{\text{a}\times16+\text{b}\times28}{\text{a}+\text{b}}=20\text{g mol}^{-1}$ (Given)
i.e. 16a + 286 = 20 (a + b) or 4a + 7b = 5 (a + b)
$\text{or }\text{a}=2\text{b}\text{ or }\frac{\text{a}}{\text{b}}=\frac{2}{1}=2:1$
If the ratio is reversed, now the ratio a : b = 1 : 2
$\therefore$ Average molar mass $=\frac{1\times16+2\times28}{1+2}=24\text{g mol}^{-1}$
View full question & answer→Question 83 Marks
If 2 litres of $\mathrm{N}_2$ is mixed with 2 litres of $\mathrm{H}_2$ at a constant temperature and pressure, then what will be the volume of $\mathrm{NH}_3$ formed?
Answer$\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_2(\mathrm{~g})$
1 L of $\mathrm{N}_2$ will react with 3 L of $\mathrm{H}_2$
L of $\mathrm{N}_2$ will react with 6 L of $\mathrm{H}_2$ but we have only
2 L of $\mathrm{H}_2$ therefore, $\mathrm{H}_2$ is limiting reactant.
3 L of $\mathrm{H}_2$ gives 2 L of NH
$\Rightarrow 2 \mathrm{~L}$ of $\mathrm{H}_2$ gives $\frac{2}{3} \times 2=\frac{4}{3}$
$=1.33 \mathrm{~L}$ of $\mathrm{NH}_3$
View full question & answer→Question 93 Marks
Calculate the weight of FeO formed from 2 g of VO and 5.75 g of $Fe _2 O _3$. Also, report the limiting reagent.
$2 VO+3 Fe_2 O 3 \rightarrow 6 FeO+V_2 O_5$
(Atomic mass of $V =51.4, O =16, Fe =55.9 g$ )
Answer$2VO + 3Fe_2O_3 → 6FeO + V_2O_5$
Molar mass of $VO= 51.4 + 16 = 67.4$
Molar mass of $Fe_2O_3 = 2 \times 55.9 + 3 \times 16$
$= 111.8 + 48 = 159.8$
$2 \times 67$.4g of $VO$ reacts with $3 \times 159$.8g of $Fe_2O_3$
2g of VO reacts with $\frac{3\times159.8}{2\times67.4}\times2=\frac{479.4}{67.4}$
$=7.1\text{g}\text{ of Fe}_2\text{O}_3$
Since $Fe _2 O _3$ is present in small amount, therefore, it is limiting reactant.
$3 \times 159.8 g$ of $Fe _2 O _3$ gives $6(55.9+16) g$ of FeO
5.75 g of $Fe _2 O _3$ gives $\frac{6 \times 71.9}{3 \times 159.8} \times 5.75$
$=5.174 g$ of FeO
View full question & answer→Question 103 Marks
How many grams of $KClO_3$ must be decomposed to prepare 3.36 litres of oxygen at STP? (Atomic weight of K = 39, Cl= 35.5, 0 = 16u)
Answer$2\text{KClO}_3(\text{s})\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ }2\text{KCl} \ \ \ \ +\ \ \ \ \ \ 3\text{O}_2\$2\times122.5=245\text{g})\ \ \ \ \ \ \ \ \ \ \ \ \ (3\times22.4\text{L})$
Since, 3 × 22.4 L of O, at STP is liberated from 245g of $KClO_3$
Therefore, 3.36 L of O, at STP is liberated from
$\frac{245}{3\times22.4}\times3.36$
$=\frac{823.2}{67.2}=12.25\text{g of KClO}_3$
View full question & answer→Question 113 Marks
Calculate the number of atoms present in 1.4 g of $\mathrm{N}_2$ molecule.
Answer28 g of $\mathrm{N}_2$ molecules contain $2 \times 6.022 \times 10^{23}$ atoms
Hence, 1.4 g of $\mathrm{N}_2$ molecules contain
$\frac{2\times6.022\times10^{23}}{28}\times1.4$
$=6.022\times10^{22}\text{ atoms}.$
View full question & answer→Question 123 Marks
If 2L of $N_2$ is mixed with 2L of H at a constant temperature and pressure, then what will be the volume of $NH_3$ formed?
Answer$\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{~g})$
1 L of $\mathrm{N}_2$ reacts with 3 L of $\mathrm{H}_2$.
Therefore, 2 L of $\mathrm{N}_2$ will react with 6 L of $\mathrm{H}_2$ but we have only 2 L of $\mathrm{H}_2$,therefore, $\mathrm{H}_2$ is the limiting reactant. 3 L of $\mathrm{H}_2$ gives 2 L of $\mathrm{NH}_3$.
$\therefore2\text{L}\text{ of H}_2\text{ gives}=\frac{2}{3}\times2=\frac{4}{3}$
$=1.33\text{L of NH}_3$
View full question & answer→Question 133 Marks
Compute the mass of one molecule and the molecular mass of $\mathrm{C}_6 \mathrm{H}_6$ (benzene) (At. mass of $\mathrm{C}=12, \mathrm{H}=1 \mathrm{u}$ ).
AnswerMolecular weight of $\mathrm{C}_6 \mathrm{H}_6=6 \times 12+6 \times 1=78 \mathrm{~g}$
Moass of 1 molecule $=\frac{78}{6.022\times10^{23}}\text{g}$
$=12.94\times10^{-23}\text{g}$
$=1.294\times10^{-22}\text{g}$
View full question & answer→Question 143 Marks
A black dot used as a full stop at the end of a sentence has a mass of about one attogram. Assuming that the dot is made up of carbon, calculate the approximate number of carbon atoms present in the dot?
AnswerMass of carbon in the dot = 1 attogram = $10^{-18}gGram$ atomic mass of carbon = 12g,
i.e. $12g$ of carbon contains $6.022 \times 10^{23}$ atoms of carbon.
$\therefore 10^{-18}g$ of carbon will contain carbon atoms
$=\frac{6.022\times10^{23}}{12}\times10^{-18}$
$=5.02\times10^{4}\text{ atoms}.$
View full question & answer→Question 153 Marks
Chlorophyll, the green colouring matter of plants contains $2.68\%$ magnesium by weight. Calculate the number of magnesium atoms in $2.00g$ of chlorophyll (Atomic mass of Mg = $24$).
AnswerMass of chlorophyll $=2.0 \mathrm{~g}$
Percentage of $\mathrm{Mg}=2.68 \mathrm{~g}$
Mass of Mg in 2.0 g of chlorophyll $=\frac{2.68 \times 2.0}{100}=0.054 \mathrm{~g}$
$6.022 \times 10^{23}$ atoms of magnesium $=24 \mathrm{~g}$
$\therefore 24 \mathrm{~g}$ of Mg contains $6.022 \times 10^{23}$ atoms
$\therefore 0.054 \mathrm{~g}$ of Mg contains $=\frac{6.022 \times 10^{23}}{24}=0.054$
$=1.3 \times 10^{21}$ atoms.
View full question & answer→Question 163 Marks
One mole of any substance contains $6.022 \times 10^{22}$ atoms/ molecules. Calculate the number of molecules of $\mathrm{H}_2 \mathrm{SO}_4$ present in 100 mL of $0.02 \mathrm{M} \mathrm{~H}_2 \mathrm{SO}_4$ Solution.
AnswerNo. of milli moles $=\mathrm{M} \times$ Volume in mL
No. of molecules $=$ No. of moles $\times 6.022 \times 10^{23}$
$=\frac{100\times0.02}{1000}\times6.022\times10^{23}$
$=12.044\times10^{20}$ molecules.
View full question & answer→Question 173 Marks
How much sugar $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)$ will be required if each person on the earth is given $10^{15}$ moles of sugar per day. Population of the earth is $3 \times 10^{10}$.
AnswerAmount of sugar $=3 \times 10^{10} \times 10^{15}=3 \times 10^{25} \mathrm{~mol}$
$=3 \times 10^{25} \times 342 \mathrm{~g}=1026 \times 10^{25} \mathrm{~mol}$
That is the amount of sugar $=1.026 \times 10^{28} \mathrm{~g}$.
View full question & answer→Question 183 Marks
If $4g$ of $NaOH$ dissolves in 36g of $H_2O$, calculate the mole fraction of each component in the solution. Also, determine the molarity of solution (specific gravity of solution is $1g mL^{–1}).$
AnswerMole fraction of $H_2O$
$=\frac{\text{No. of moles of H}_2\text{O}}{\text{No. of moles}(\text{H}_2\text{O}+\text{NaOH})}$
No. of moles of $H_2O$ $=\frac{36}{18}=2\ \text{moles}$
No. of moles NaOH $=\frac{4}{40}=0.1\ \text{mol}$
Total number of moles = 2 + 0.1 = 2.1
Mole fraction of $H_2O$$=\frac{2}{2.1}=0.952$
Mole fraction of NaOH $=\frac{0.1}{2.1}=0.48\text{b}$ Mass of solution
= Mass of $H_2O$ + Mass of NaOH = 36 + 4 = 40g
Volume of solution $=\frac{40}{1}=40\text{mL}$
Molarity = No.of moles of solute/ Volume of solution in L
$=\frac{0.1}{0.4}\text{L}=2.5\text{M}$
View full question & answer→Question 193 Marks
The water level in a metric measuring cup is 0.75L before the addition of a pebble weighing 150g. The water level after submerging the pebble is 0.82L. Determine the density of the pebble.
AnswerThe volume displaced by the pebble
= 0.82 - 0.75 = 0.07L = 70mL
Mass of the pebble = 150g
Therefore, density of the pebble is the
$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$
$=\frac{150}{70}=2.14\text{g mL}^{-1}$
View full question & answer→Question 203 Marks
The density of 3 molal solution of NaOH is $1.110 \mathrm{~g} \mathrm{~mL}^{-1}$. Calculate the molarity of the solution.
Answer3 molal solutions of NaOH means 3 moles of NaOH is dissolved in 1000 g of water.
But 3 moles of $\mathrm{NaOH}=3 \times 40 \mathrm{~g}=120 \mathrm{~g}$
$120 \mathrm{~g}=1120 \mathrm{~g}$
Density os solution $=1.110\text{g}\ \text{mL}^{-1}$
$\therefore$ volume of solution $=\frac{\text{mass}}{\text{Density}}=\frac{1120\text{g}}{1.11\text{g}\ \text{L}^{-1}}=1009\text{mL}=1.009\text{L}$
Molarity of solution $=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}=\frac{3\text{mple}}{1.009\ \text{L}}=2.97\ \text{M}$
View full question & answer→Question 213 Marks
What is the molecular mass of a substance each molecule of which contains 9 atoms of carbon, 13 atoms of hydrogen and $2.33 \times 10^{-23} \mathrm{~g}$ other component?
AnswerMass of 9 atoms of carbon $=9 \times 12 \mathrm{amu}=108 \mathrm{u}$ Mass of 13 atoms of hydrogen $=13 \times 1 \mathrm{amu}=13 \mathrm{u}$ Mass of $2.33 \times$ $10^{-23} \mathrm{~g}$ of other component $=(1 \mathrm{u}) \times \frac{\left(2.33 \times 10^{-23} \mathrm{~g}\right)}{\left(1.66 \times 10^{-24} \mathrm{~g}\right)}=14.04 \mathrm{uMolecular}$ mass of the substance $(108+13+14.04) u=135.04 u$.
View full question & answer→Question 223 Marks
Calculate the mass of ferric oxide that will be obtained by complete oxidation of 2g of Fe. [Atomic weights of Fe = 56u, 0 = 16u]
Answer$4 \mathrm{Fe}+3 \mathrm{O}_2 \longrightarrow 2 \mathrm{Fe}_2 \mathrm{O}_3 4 \times 56 \mathrm{~g}$ Fe gives $2 \times 160 \mathrm{~g}$ of $\mathrm{Fe}_2 \mathrm{O}_3$.
2 g of Fe gives $\frac{2 \times 160}{4 \times 56} \times 2=2.857 \mathrm{~g}$ of $\mathrm{Fe}_2 \mathrm{O}_3$.Molecular weight of $\mathrm{Fe}_2 \mathrm{O}_3=2 \mathrm{Fe}+3 \mathrm{O}$ $=2 \times 56+3 \times 16=112+48=160 \mathrm{~g} \mathrm{~mol}^{-1}$.
View full question & answer→Question 233 Marks
Calculate the concentration of nitric acid in mol per litre in a sample which has a density $1.41g mL^{-1}$ and the mass percent of nitric acid in it being $69\%.$
AnswerMolarity $=\frac{\text{w}\times1000}{\text{m}\times\text{Volume of solution (mL)}}$
Given, $d =1.41 g mL ^{-1}$, mass $\%$ of $HNO _3=69 \%$
$69 \% HNO _3$ means 100 g of its solution contains $69 g HNO _3$ (nitric acid).
Hence, mass of $HNO _3$ (solute) $=69 g$
Molar mass of nitric acid,
$HNO_3=1.0079+14.0067+(3 \times 16.00)=63.0146 g mol^{-1}$
Density, $\text{d}=\frac{\text{m}}{\text{V}}\text{ or }\text{V}=\frac{\text{m}}{\text{d}}$
$=\frac{100\text{g}}{1.41\text{g mL}}^{-1}$
Molarity $=\frac{\text{w}\times1000}{\text{m}\times\text{Volume of solution (mL)}}$
$=\frac{69\times1000\times1.41}{63.0146\times100}=15.44\text{M}.$
Note: Concentration of a substance in mol per litre is known as molarity.
View full question & answer→Question 243 Marks
Calculate the mass of $112 \mathrm{~cm}^3$ of hydrogen gas at STP.
(Atomic mass of $\mathrm{H}=1 \mathrm{u}$ )
AnswerSince $22400 \mathrm{~cm}^3$ of hydrogen at STP weighs $=2 \mathrm{~g}$
Therefore, $112 \mathrm{~cm}^3$ of hydrogen at STP weighs
$=\frac{2}{22400} \times 112=0.01 \mathrm{~g} .$
View full question & answer→Question 253 Marks
If $6.022 \times 10^{23}$ molecules of $\mathrm{N}_2$, react completely with $\mathrm{H}_2$ according to the equation:
$\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{~g})$
then calculate the number of molecules of $\mathrm{NH}_3$ formed.
Answer$\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{~g})$
$6.022 \times 10^{23}$ molecules of $\mathrm{N}_2$, react completely with $\mathrm{H}_2$ to give $2 \times 6.023 \times 10^{23}$ molecules of $\mathrm{NH}_3=1.204 \times 10^{24}$ molecules.
View full question & answer→Question 263 Marks
56 kg of $\mathrm{N}_2(\mathrm{~g})$ and 10 kg of $\mathrm{H}_2(\mathrm{~g})$ are mixed to produce $\mathrm{NH}_3(\mathrm{~g})$. Calculate the number of moles of ammonia gas formed.
(Atomic mass/ $\mathrm{g} \mathrm{mol}^{-1} \mathrm{~N}=14, \mathrm{H}=1$ )
Answer$\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})$
Since 28 kg of $\mathrm{N}_2$ reacts with 6 kg of $\mathrm{H}_2$.
Therefore, 56 kg of $\mathrm{N}_2$ reacts with $\frac{6}{28} \times 56=12 \mathrm{~kg}$ of $\mathrm{H}_2$.
But we have only 10 kg of $\mathrm{H}_2$, therefore, $\mathrm{H}_2$ is limiting reactant.
Also, 6 kg of $\mathrm{H}_2$ will give 2 moles of $\mathrm{NH}_3$.
Hence, 10 kg of $\mathrm{H}_2$, will give $\frac{2}{6} \times 10=3.33$ moles of $\mathrm{NH}_3$.
View full question & answer→Question 273 Marks
1 M solution of $\mathrm{NaNO}_3$ has density $1.25 \mathrm{~g} \mathrm{~cm}^{-3}$. Calculate its molality.
(Mol. weight of $\mathrm{NaNO}_3=85 \mathrm{~g} \mathrm{~mol}^{-1}$ )
AnswerMass of solution $=$ Volume of solution $\times$ Density of solution
Mass of solution $=1000 \mathrm{~cm}^3 \times 1.25 \mathrm{~g} \mathrm{~cm}^{-3}=1250 \mathrm{~g}$
Mass of solute $=85 \mathrm{~g}$
Mass of solvent $=1250-85=1165 \mathrm{~g}$
$\text{M}=\frac{\text{W}_{\text{B}}}{\text{M}_{\text{B}}}\times\frac{1000}{\text{W}_{\text{A}}\text{in grams}}$
$=\frac{85}{85}\times\frac{1000}{1165}=0.858\text{m}.$
View full question & answer→Question 283 Marks
Calculate the total number of electrons present in $1.6g$ of methane.
Answer
- Molar mass of methane $(CH_4)= 12 + 4 \times 1 = 16g$
16g of methane contains $= 6.022 \times 10^{23}$ molecules
1.6g of methane will contain
$=\frac{7.022\times10^{23}}{(16\text{g})}\times(1.6\text{g})$
$=6.022\times10^{22}$ molecules
- Number of electrons in $6.022 \times 10^{22}$ molecules of methane. 1 molecule of methane contains $=6+4=10$ is electrons $6.022 \times 10^{22}$ molecules of methane contain electrons $=6.022 \times 10^{22} \times 10=6.022 \times 10^{23}$.
View full question & answer→Question 293 Marks
Balance the following equations:
- $H_3PO3 → H_3PO_4+ PH_3$
- $Ca + H_2O → Ca(OH)_2 +H_2$
- $Fe_2(SO_4)_3+ NH_3 + H_2O → Fe(OH)_3 + (NH_4)_2SO_4$
- $Cl_2+ NaOH → NaCl + NaClO_3 + H_2O$
Answer
- $4H_3PO_3→ 3H_3PO_4+ PH_3$
- $Ca + 2H_2O → Ca(OH)_2 +H_2$
- $Fe_2(SO_4)_3+ 6NH_3 + 6H_2O → 2Fe(OH)_3 + 3(NH_4)_2SO_4$
- $3\text{Cl}_2+ 6\text{NaOH}\xrightarrow{\text{heat}}5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}$
View full question & answer→Question 303 Marks
If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in whole number ratio.
- Is this statement true?
- If yes, according to which law?
- Give one example related to this law.
Answer
- Yes, the statement is true.
- According to law of multiple proportions.
- Hydrogen and oxygen react to produce two compounds, water and hydrogen peroxide. Masses of oxygen which combine with fixed mass of hydrogen are in simple ratio.
$\text{H}_2+\frac{1}{2}\text{O}_2\rightarrow\text{H}_2\text{O}\\^{2\text{g}}\ \ \ \ \ \ \ ^{16\text{g}}\ \ \ \ \ \ \ ^{18\text{g}}$ View full question & answer→Question 313 Marks
- How many gram atoms are there in $8.0g$ of S?
- The molarity of Solution of sulphuric acid is $1.35M$. Calculate its molality.
(The density of solution is $1.02g cm^{-3})$Answer
32g of S = 1g atom
$\Rightarrow8\text{g of S}=\frac{1}{32}\times8=0.25\text{g atom}$
$M = 1.35mol L^{-1}$
Mass of solution = Volume of solution × Density of solution
Mass of solute = M × Molecular mass of $H_2SO_4$
$= 1.35 \times 98g = 132.3g$
Mass of solvent = Mass of solution - Mass of solute
$= 1020g - 132.3g = 887.7g$
Molality $=\frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}}$
$=\frac{1.35\text{m}}{\frac{88.7}{1000}\text{kg}}=\frac{1350}{887.7}=1.52\text{mol/ kg}.$
View full question & answer→Question 323 Marks
Express each of the following in SI units,
- 14 pound per square inch (atmospheric pressure).
- 100 mile per hour.
- 5 feet 2 inch.
- 150 pound.
Answer
- 1 pound per square inch = $6894.76Nm^{-2}$
$\therefore$ 14 pound per square inch
$= 14 × 6894.76 = 96526.64Nm^{-2}$
- $1\text{mile/h}=\frac{1.6\times10^{3}\text{m}}{3600\text{s}}=0.444\text{ms}^{-1}$
- 5 feet 2 inch = 1.5748m
($\therefore$ 1 feet = 12 inch, 1m = 39.37 inch]
- 1 pound = 0.454kg
$\therefore$ 150 pound = 68.1kg. View full question & answer→Question 333 Marks
How many atoms are present in 1 ml of $\mathrm{NH}_3$ at STP?
Answer$22400 \mathrm{ml} \text { of } \mathrm{NH}_3 \text { contains }=4 \times 6.022 \times 1023 \text { atoms }$
${\left[\because \mathrm{NH}_3 \text { contains } 4 \text { atoms }\right]}$
$1 \mathrm{ml} \text { of } \mathrm{NH}_3 \text { contains }=\frac{4 \times 6.022 \times 10^{23}}{22400}$
$
$=1.07 \times 10^{20} \text { atoms. }$
View full question & answer→Question 343 Marks
Calcium carbonate reacts with aqueous HCl to give $CaCl _2$ and $CO _2$ according to the reaction given below:
$CaCO_3(s)+2 HCl(aq) \rightarrow CaCl_2(aq)+CO_2(g)+H_2 O(l)$
What mass of $CaCl _2$ will be formed when 250 mL of 0.76 M HCl reacts with 1000 g of $CaCO _3$ ? Name the limiting reagent. Calculate the number of moles of $CaCl _2$ formed in the reaction.
AnswerNumber of moles of HCl takan $=\frac{\text{MV}}{1000}=\frac{0.76\times250}{1000}=0.19$
Number of moles of $CaCO_3$ $=\frac{\text{Mass}}{\text{Molar mass}}=\frac{1000}{100}=10$
The given reaction is:
$\text{CaCo}_3+2\text{HCl}\rightarrow\text{CaCl}_2+\text{CO}_2+\text{H}_2\text{O}\\ \ \ \ \ ^{1\text{mol}}\ \ \ \ \ \ \ ^{2\text{mol}}\ \ \ \ \ \ \ \ \ \ \ \ ^{1\text{mol}}$
Case I: Let $CaCO _3$ is completely consumed.
$1 mol CaCO_3=1 mol CaCl_2$
$\therefore 10 mol CaCO_3=10 mol CaCl_2$
Case II: Let HCl is completely consumed.
$2 mol HCl=1 mol CaCl_2$
$0.19 mol HCl=\frac{1}{2} \times 0.19 mol CaCl_2=0.095 mol CaCl_2$
Since HCl on complete consumption gives lesser amount of product hence HCl will be limiting reagent and the number of moles of $CaCl _2$ formed will be 0.095 mol .
View full question & answer→Question 353 Marks
An organometallic compound on analysis was found to contain, C = 64.4%, H = 5.5% and Fe= 29.9%. Determine its empirical formula (At. mass of Fe = 56u).
Answer
| Element |
% |
Atomic mass |
Relative number of moles |
Simplest molar ratio |
Simplest whole number molar Ratio |
| C |
64.4 |
12 |
$\frac{64.4}{12}=5.36$ |
$\frac{5.36}{0.53}=10.1$ |
10 |
| H |
5.50 |
1 |
$\frac{5.50}{1}=5.50$ |
$\frac{5.50}{0.53}=10.4$ |
10 |
| Fe |
29.9 |
56 |
$\frac{29.9}{56}=0.53$ |
$\frac{0.53}{0.53}=1$ |
1 |
Therefore, empirical formula= $\mathrm{C}_{10} \mathrm{H}_{10} \mathrm{Fe}$. View full question & answer→Question 363 Marks
0.5 mole each of $\mathrm{H}_2 \mathrm{~S}$ and $\mathrm{SO}_2$ mixed together in a reaction Flask, react according to equation:
$2 \mathrm{H}_2 \mathrm{~S}+\mathrm{SO}_2 \rightarrow 2 \mathrm{H}_2+3 \mathrm{~S}$
Calculate the number of moles of 'S' formed.
Answer$2 \mathrm{H}_2 \mathrm{~S}+\mathrm{SO}_2 \rightarrow 2 \mathrm{H}_2+3 \mathrm{~S}$
Now, 2 moles of $\mathrm{H}_2 \mathrm{~S}$ combine with 1 mole of $\mathrm{SO}_2$ to give 3 moles of S .
1 mole of $\mathrm{H}_2 \mathrm{~S}$ combines with 0.5 mole of $\mathrm{SO}_2$ to give $\frac{3}{2}$ moles of S .
Therefore, 0.5 mole of $\mathrm{H}_2 \mathrm{~S}$ combines with 0.25 mole of $\mathrm{SO}_2$ to give $\frac{3}{2} \times 0.5=0.75$ moles of S .
View full question & answer→Question 373 Marks
Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc. Following reaction takes place.
$Zn + 2HCl → ZnCl_2 + H_2$
Calculate the volume of hydrogen gas liberated at STP when 32.65g of zinc reacts with HCl. 1mol of a gas occupies 22.7L volume at STP; atomic mass of $Zn = 65.3u.$
AnswerGiven that, mass of Zn = 32.65g1 mole of gas occupies = 22.7L volume at STP Atomic mass of Zn = 65.3u The given equation is
$\text{Zn}+2\text{HCl}\rightarrow\text{ZnCl}_2+\text{H}_2\\^{65.3\text{g}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{1\text{mol=22.7}\text{Lat}\text{STP}}$
From the above equation, it is clear thet 65.3g of Zn when reacts with HCl produce
$=22.7\ \text{L H}_2$ at STP
$\therefore$ 32.65g of Zn when reacts with HCl will produce $=\frac{22.7\times32.65}{65.3}=11.35\text{L}$ of $H_2$ at STP
View full question & answer→Question 383 Marks
A mixture of gases containing $\text{H}_2$ and $\text{O}_2$ in the ratio of 1 : 4 by mass. What is their molar ratio.
AnswerNumber of moles $=\frac{\text{Mass}}{\text{Molar mass}}$
Number of moles of $\text{H}_2=\frac{1}{2}=\frac{1}{2}$
Number of moles of $\text{O}_2=\frac{4}{32}=\frac{1}{8}$
Ration $=\frac{1}{2}:\frac{1}{8}=4:1$
View full question & answer→Question 393 Marks
Calculate the concentration of HBr solution in $\mathrm{mol} \mathrm{~L}^{-1}$ in a sample which has density $1.5 \mathrm{ml}^{-1}$ and mass percent of HBr being $48 \%$. [Molar mass of $\left.\mathrm{HBr}=81 \mathrm{g} \mathrm{~mol}^{-1}\right]$
AnswerMass of solute = 48g, Mass of solution = 100g,
Volume of solution $=\frac{\text{M}}{\text{d}}=\frac{100}{1.5}$
$\text{M}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}}\times\frac{1000}{\text{Volume of solution}}$
$=\frac{48\times1000}{81\times\frac{100}{1.5}}=8.8\text{mol L}^{-1}$
View full question & answer→Question 403 Marks
A vessel contains 1.6g of dioxygen at STP ($273.15K, 1$ atm pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes half of the original pressure. Calculate
- Volume of the new vessel.
- Number of molecules of dioxygen.
Answer
$$Moles of oxygan $=\frac{1.6}{32}$ = 0.05mol
1mol of $O_2$ at STP has volume = 22.4L
0.05mol of $O_2$ at has vollume $= 22.4 \times 0.05 = 1.12L$
$V_1 = 1.12L$
$P_1 = 1 atm$
$V_2 = ?$
$P_2$ $=\frac{1}{2}$ = 0.5 atm
According to Boyle's law (unit 4)
$P_1 V_1 = P_2 V_2$
or $\text{V}_2=\frac{\text{p}_1\text{V}_1}{\text{p}_2}=\frac{1\text{atm}\times1.12\text{L}}{0.5\ \text{atm}}=2.24\text{L}$
No. of molecules in 1.6g or 0.05mol
$= 6.022 \times 10^{23} \times 0.05 = 3.011 \times 10^{22}.$ View full question & answer→Question 413 Marks
Calculate the percentage of copper in a sample of $\mathrm{CuCl}_2$.
(Atomic mass of $\mathrm{Cu}=63.5 \mathrm{u}, \mathrm{Cl}=35.5 \mathrm{u}$ )
AnswerMolecular mass of $\mathrm{CuCl}_2=63.5+2 \times 35.5$
$=63.5+71=134.5 u$
$\%\text{of Cu}=\frac{\text{Total mass of copper}}{\text{molecular mass of CuCl}_2}\times100$
$=\frac{63.5}{134.5}\times100=47.21\%$
View full question & answer→Question 423 Marks
What is the concentration of sugar $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right) \mathrm{i} \mathrm{~mol} \mathrm{~L}^{-1}$ if its 20 g are dissolved in enough water to make a final volume up to 2 L ?
AnswerMolar mass of the sugar, $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$
$\mathrm{m}=(12 \times 12.01)+(22 \times 1.0079)+(11 \times 16.00)$
$=342.2938 \mathrm{~g} \mathrm{~mol}^{-1}=342$
Given, $\mathrm{w}=20 \mathrm{~g}, \mathrm{~V}=2 \mathrm{~L}$
$\text { Molarity }=\frac{\mathrm{w}}{\mathrm{~m} \times \mathrm{V}(\mathrm{~L})}=\frac{20}{342 \times 2}$
$=0.0292^{-1}=0.0292 \mathrm{M}$.
View full question & answer→Question 433 Marks
45.4L of dinitrogen reacted with 22.7L of dioxygen and 45.4L of nitrous oxide was formed. The reaction is given below:
$2 \mathrm{N}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{N}_2 \mathrm{O}(\mathrm{g})$
Which law is being obeyed in this experiment? Write the statement of the law?
AnswerGay Lussac’s Law of Gaseous Volumes is followed here. This law was given by Gay Lussac in 1808. He observed that when gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at same temperature and pressure.
45.4L of dinitrogen reacted with 22.7L of dioxygen and 45.4L of nitrous oxide.
Thus, the volumes of dinitrogen and oxygen which combine together (i.e. 45.4L and 22.7L) bear a simple ratio of 2 : 1.
Gay-Lussac’s discovery of integer ratio in volume relationship is actually the law of definite proportions by volume. The law of definite proportions is with respect to mass.
View full question & answer→Question 443 Marks
10 mL of H , combine with 5 mL of $\mathrm{O}_2$ to form water. When 200 mL of H , at STP is passed over heated CuO, the CuO loses 0.144 g of its weight. Does the above data correspond to the law of constant composition?
AnswerIn the second experiment 0.144 g weight is lost from CuO . This is due to the reduction of CuO into Cu . In other words, 0.144 g oxygen combined with $200 \mathrm{~mL} \mathrm{H}_2$.
32 g oxygen occupies 22400 mL volume at STP.
0.144 g oxygen will occupy $=22400 \times \frac{0.144}{32}$
$=100.8 \mathrm{~mL} \mathrm{O}{ }_2$
It means the ratio of $\mathrm{H}_2$ and $\mathrm{O}_2$ in water is $200: 100.8=2: 1$. The same ratio is in first case ( $10: 5$ or $2: 1$ ). Thus, the law of constant composition is proved.
View full question & answer→Question 453 Marks
- A compound contains 21.6% sodium, 33.3% chlorine, 45.1 % oxygen. Derive its empirical formula.
- Write the empirical formulae for the following compounds:
- State the number of significant figures in each of the following numbers:
Answer
-
| Element |
Percentage composition
|
Relative number of moles
|
Simplest molar ratio
|
Simplest whole number molar ratio
|
|
Na
|
$21.6$
|
$0.939$
|
$1$
|
$1$
|
|
Cl
|
$33.3$
|
$0.938$
|
$1$
|
$1$
|
|
O
|
$45.1$
|
$2.82$
|
$3$
|
$3$
|
Hence, the empirical formula in $NaClO_3$
-
- $H_2O$
- $BH_3$
-
View full question & answer→Question 463 Marks
5.975g of the higher oxide of metal gave 5.575g of lower oxide on heating. The quantity of the lower oxide gave 5.175g of metal on reduction. Prove that these results are in accordance with the law of multiple proportions.
AnswerAs 5.575g of lower oxide on reduction gives 5.175g of the metal, the mass of the oxygen is 5.575 - 5.175 = 0.4g
For 1g of metal, mass of oxygen is $\frac{0.4}{5.175}= 0.077\text{g}$
In case of higher oxides, mass of metal is 5.175g
Mass of oxygen is 5.975 - 5.175 = 0.8g
For 1g of metal, mass of oxygen is $\frac{0.8}{5.175}= 0.155\text{g}$
For a given mass of metal, the ratio of oxygen is 0.777: 0.155 or 1:2. Hence, the law of multiple proportion is proved.
View full question & answer→Question 473 Marks
Chlorine is prepared in the laboratory by treating manganese dioxide ($MnO_2$) with aqueous hydrochloric acid according to the reaction
$4HCl_{(aq)} + MnO_{2(s)} → 2H_2O_{(l)} + MnCl_{2(aq)} + Cl_{2(g)}$
How many grams of HCl react with 5.0g of manganese dioxide?
Answer$1mol [55 + 2 \times 16 = 87g] MnO_2$ reacts completely with 4mol $[4 \times 36.5 = 146g]$ of HCl.
$\therefore$ 5.0g of $MnO_2$ will react with
$=\frac{146\text{g}}{87\text{g}}\times5.0\text{g}$ of HCl
$= 8.4g$ of $HCl$
Hence, $8.4g$ of HCl will react completely with 5.0g of manganese dioxide.
View full question & answer→Question 483 Marks
Commercially available concentrated hydrochloric acid contains $38 \% \mathrm{HCl}$ by mass.
i. What is the molarity of the solution (density of solution $=1.19 \mathrm{~g} \mathrm{~mL}^{-1}$ )?
ii. What volume of the above concentrated HCl is required to make 1.0 L of 0.10 M HCl ?
Answer
- 38% HCl by mass means 38g of HCl is present in 100g of solution.
Volume of solution $=\frac{\text{mass}}{\text{density}}=\frac{100}{1.19}=84.03\text{mL}$
Moles of HCl $=\frac{38}{36.5}=1.04$
Molarity $=\frac{1.04\times1000}{84.03}=12.38\text{M.}$
- From the Molarity equation,$\underbrace{\text{M}_1\text{V}_1}=\underbrace{\text{M}_2\text{V}_2}\\\text{acid}_1\ \ \ \ \ \ \ \ \ \text{acid}_2$
$12.38\text{M}\times\text{V}_1=0.10\text{M}\times1.0\text{L}$
$\therefore\text{V}_10=\frac{0.1\times1.0}{12.38}$
$=0.00808\text{L}=8.08\text{cm}^3$ View full question & answer→Question 493 Marks
- Calculate the mole fraction of water in a mixture of $15g$ water, $18g$ acetic acid and $84g$ ethyl alcohol.
- Excess the following numbers to four significant figures.
- $3.607892$
- $6.5869 \times 10^3$
- Write balanced chemical equation for the following:
$KMnO_4 + C_2H_4 + H_2O → MnO_2 +KOH+ (CH_2OH)_2$Answer
- $\text{n}\big(\text{H}_2\text{O}\big)=\frac{\text{Mass of water}}{\text{Molar mass of water}}$
$=\frac{15\text{g}}{18\text{g mol}^{-1}}=0.84\text{mol}$
$\text{n}\big(\text{CH}_3\text{COOH}\big)=\frac{\text{Mass of acetic acid}}{\text{Molar mass of acetic acid}}$
$=\frac{78}{60\text{g mol}^{-1}}=1.64\text{mol}$
$\text{n}\big(\text{C}_2\text{H}_5\text{OH}\big)=\frac{\text{Mass of ethyl alcohol}}{\text{Molar mass of ethyl alcohol}}$
$=\frac{84\text{g}}{46\text{g mol}^{-1}}=1.82\text{mol}$
So, Total number of moles in the solution,
n total = (0.84 + 1.82 + 1.64)mol = 4.30mol
Therefore, $\text{X}_{{\text{water}}}=\frac{\text{n}\big(\text{H}_2\text{O}\big)}{\text{n}_{\text{total}}}$
$=\frac{1.64}{4.50}=0.38$
$\text{X}_{{\text{acetic acid}}}=\frac{\text{n}\big(\text{CH}_3\text{COOH}\big)}{\text{n}_{\text{total}}}$
$=\frac{0.84\text{mol}}{4.30\text{mol}}=0.19$
$\text{X}_{\text{ethyl alcohol}}=\frac{\text{n}\big(\text{C}_2\text{H}_5\text{OH}\big)}{\text{n}_{\text{total}}}$
$=\frac{1.82\text{mol}}{4.30\text{mol}}=0.42$
-
- $2KMnO_4 +3C_2H_4 + 4H_2O → 2 MnO_2 + 2KOH + 3(CH_2OH)_2$
View full question & answer→Question 503 Marks
For precious stone, carat is used for specifying its mass. If 1 carat $= 3.08647$ grains (a unit of mass) and $1$ gram $= 15.4324$ grains. Find the total mass in kilogram of a ring that contains $0.700$ carat diamond and 5.00 gram gold.
AnswerFinding mass of diamond in kg 1 carat = 3.08647 grains
$\therefore$ 0.700 carat = 0.700 × 3.08647 grains = 2.16 grains Also, 1 gram = 15.4324 grains
$\therefore1\text{grain}=\frac{1}{15.4324}=0.064799\text{ grams}.$and 2.16 grains = 216 × 0.064799 = 0.1399 grams
$1$ gram $= 10^{-3}/ g$
$\Rightarrow 0.1399g = 1.399 \times 10^{-2}kg$ Total mass of the ring in kg $= 1.399 \times 10^{-2} kg + 5 \times 10^{-3}kg = 1.899 \times 10^{-2}kg.$
View full question & answer→Question 513 Marks
A flask P contains 0.5 mole of oxygen gas. Another flask Q contains 0.4 mole of ozone gas. Which of the two flasks contain greater number of oxygen atoms?
Answer1 molecule of oxygen $\left(\mathrm{O}_2\right)=2$ atoms of oxygen
1 molecule of ozone $\left(\mathrm{O}_3\right)=3$ atoms of oxygen
In flask P, 1 mole of oxygen gas $=6.022 \times 10^{23}$ molecules
$\therefore 0.5$ mole of oxygen gas $=6.022 \times 10^{23} \times 0.5$ molecules
$=6.022 \times 10^{23} \times 0.5 \times 2$ atoms $=6.022 \times 10^{23}$ atoms
In flask Q 1 mole of ozone gas $=6.022 \times 10^{23}$ molecules
0.4 mole of ozone gas $=6.022 \times 10^{23} \times 0.4$ molecules
$=6.022 \times 10^{23} \times 0.4 \times 3$ atoms $=7.23 \times 10^{23}$ atoms
$\therefore$ Flask $Q$ has greater number of oxygen atoms as compared to the flask $P$.
View full question & answer→Question 523 Marks
A mixture of oxalic acid and formic acid is heated with concentrated $H_2SO_4$ and the gas evolved is collected. On treating the solution with KOH, the volume of the solution decreases by $\frac{1}{6}^\text{th}$ Calculate the molar ratios of two acids in the original mixture.
AnswerLet x mole of oxalic acid and y mole of formic acid be heated with conc. $H_2SO_4$ according to the following equations. $\text{COOH}-\text{COOH}\xrightarrow{\text{H}_2\text{SO}_4/{\text{heat}}}\text{CO(g)}+\text{CO}_2\text{(g)}+\text{H}_2\text{O(l)}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{x mol} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{x mol}\ \ \ \ \ \ \ \text{x mol}$ $\text{HCOOH}\xrightarrow{\text{H}_2\text{SO}_4/{\text{heat}}}\text{CO(g)}+\text{H}_2\text{O(l)}\\\text{y mol}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{y mol}$ Total moles of gaseous mixture = Moles of CO + Moles of $CO_2$ = (x + y) mol + x mol = (2x + y)mol Now, KOH absorbs only Co, i.e. x moles and the volume of the solution decreases by $\frac{1}{6}^\text{th}$ of its volume. Since equal volume of gases have equal number of moles according to Avogadro's law,$\therefore\frac{\text{moles of CO}_2}{\text{moles of both the gases}}$
$=\frac{\text{x}}{(2\text{x}+\text{y})}=\frac{1}{6}$ $\Rightarrow 6\text{x}=2\text{x}+\text{y}$ $\Rightarrow 4\text{x}=\text{y}$ $\Rightarrow \frac{\text{y}}{\text{x}}=4$ $\therefore$ Molar ratio of formic acid : oxalic acid = 4 : 1.
View full question & answer→Question 533 Marks
Describe what you need to do in the laboratory to test (i) the law of conservation of mass, (ii) the law of definite proportion and (iii) the law of multiple proportions.
Answer
- To test the law of conservation of mass, a reaction would have to be carried out in which the mass of the reactants and the mass of the products are weighed and shown to be the same.
- The law of definite proportions could be shown by demonstrating that no matter how a compound is obtained, the reactants remain at the same proportions by mass. This can be done by decomposing a compound and showing that the masses of the elements present are always in the same ratio.
- To test the law of multiple proportions, two different compounds made up of the same elements would have to be decomposed. If the mass of one of the elements is kept constant the masses of other elements combining with that of the element in different samples would have to be in the small whole number ratio.
View full question & answer→Question 543 Marks
- How can you say that sugar is solid and water is liquid?
- Oxygen is prepared by catalytic decomposition of potassium chlorate ($KClO_3$). Decomposition of potassium chlorate gives potassium chloride (KCl) and oxygen ($O_2$)· If 2.4mol of oxygen is needed for an experiment, how many grams of potassium chlorate must be decomposed?
(At. mass of K = 39, Cl = 35.5, O = 16)AnswerSugar has fixed shape and volume where as water has fixed volume but not fixed shape.
$2 KClO_3(s) \rightarrow 2 KCl(s)+3 O_2(g)$
Molar mass of $KClO _3=39+35.5+3 \times 16$
$=122.5$
For 3 moles of $O _2$, we need $2 \times 122.5 g$ of. $KClO _3$
For 2.4 moles of $O _2$, we need $=\frac{2 \times 122.5}{3} \times 2.4$
$=196 g \text { of } KClO_3$
View full question & answer→Question 553 Marks
Pressure is determined as force per unit area of the surface. The SI unit of pressure, Pascal is as shown below: $1 \mathrm{~Pa}=1 \mathrm{~N} \mathrm{~m}^{-2}$
If mass of air at sea level is $1034 \mathrm{~g} \mathrm{~cm}^{-2}$, calculate the pressure in Pascal.
AnswerPressure is defined as force acting per unit area of the surface.
$P=\frac{F}{A}$
$=\frac{1034 \mathrm{~g} \times 9.8 \mathrm{~ms}^{-2}}{\mathrm{~cm}^2} \times \frac{1 \mathrm{~kg}}{1000 \mathrm{~g}} \times \frac{(100)^2 \mathrm{~cm}^2}{1 \mathrm{~m}^2}$
$=1.01332 \times 10^5 \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-2}$
We know,
$1 \mathrm{~N}=1 \mathrm{~kg} \mathrm{~ms}^{-2}$
Then,
$1 \mathrm{~Pa}=1 \mathrm{Nm}^{-2}=1 \mathrm{~kg} \mathrm{~m}^{-2} \mathrm{~s}^{-2}$
$1 \mathrm{~Pa}=1 \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-2}$
$\therefore \text { Pressure }=1.01332 \times 10^5 \mathrm{~Pa}$
View full question & answer→Question 563 Marks
Two oxides of a metal contain 27.6% and 30.0% of oxygen respectively. If the formula of the first oxide is $\mathrm{M}_3 \mathrm{O}_4$, find that of the second.
AnswerIn the first oxide, oxygen $=27.6$
Metal $=100-27.6=72.4$ parts by mass.
As the formula of the oxide is $\mathrm{M}_3 \mathrm{O}_4$ it means
72.4 parts by mass of metal $=3$ atoms of metal and
4 atoms of oxygen $=27.6$ parts by mass.
In the second oxide, oxygen $=30.0$ parts by mass and metal $=100-30=70$ parts by mass.
But 72.4 parts by mass of metal $=3$ atoms of metal.
$\therefore 70$ parts by mass of metal $=\frac{3}{72.4} \times 70$ atoms of metal $=2.90$ atoms of metal
Also, 27.6 part by mass of oxygen $=4$ atoms of oxygen.
$\therefore 30$ part by mass of oxygen $=\frac{4}{27.6} \times 30$ atoms of oxygen
$=4.35$ atoms of oxygen.
Hence, ratio of $\mathrm{M}: \mathrm{O}$ in the second oxide $=2.90: 4.35=1: 1.5$ or $2: 3$
$\therefore$ Formula of the other metal oxide is $\mathrm{M}_2 \mathrm{O}_3$.
View full question & answer→Question 573 Marks
Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).
AnswerMole fraction of $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ $=\frac{\text{Number of moles of }\text{C}_2\text{H}_5\text{OH}}{\text{Number of moles of solution}}$
$0.040=\frac{^\text{n}\text{C}_2\text{H}_5\text{OH}}{^\text{n}\text{C}_2\text{H}_5\text{OH}+\ ^\text{n}\text{H}_2\text{O}}\ .....(1)$
Number of moles present in 1L water:
$^\text{n}\text{H}_2\text{O}=\frac{1000\text{g}}{18\text{g }\text{mol}^{-1}}$
$^\text{n}\text{H}_2\text{O}=55.55\text{mol}$
Substituting the value of $^\text{n}\text{H}_2\text{O}$ in equation (1),
$\frac{^\text{n}\text{C}_2\text{H}_5\text{OH}}{^\text{n}\text{C}_2\text{H}_5\text{OH}+55.55}=0.040$
$^\text{n}\text{C}_2\text{H}_5\text{OH}=0.040\ ^\text{n}\text{C}_2\text{H}_5\text{OH}+(0.040)(55.55)$
$0.96\ ^\text{n}\text{C}_2\text{H}_5\text{OH}=2.222\text{mol}$
$^\text{n}\text{C}_2\text{H}_5\text{OH}=\frac{2.222}{0.96}\text{mol}$
$^\text{n}\text{C}_2\text{H}_5\text{OH}=2.314\text{mol}$
$\therefore$ Molarity of solution $=\frac{2.314\text{mol}}{1\text{L}}$
$= 2.314 \text{M}$
View full question & answer→Question 583 Marks
Cone. HCl is $38 \% \mathrm{~HCl}$ by mass. What is the molarity of this solution if $\mathrm{d}=1.19 \mathrm{~g} \mathrm{~cm}^{-3}$ ? What volume of cone. HCl is required to make 1.00 L of 0.10 M HCl ?
Answer$\text{M}=\frac{\text{W}_{\text{B}}}{\text{M}_{\text{B}}}\times\frac{1000}{\text{Vol. of solution in ml}}$
$=\frac{38}{36.5}\times\frac{100}{\frac{\text{Mass of solution}}{\text{Density of solution}}}$
$=\frac{38}{36.5}\times\frac{1000}{\frac{100}{1.19}}$
$=\frac{38}{36.5}\times\frac{1000\times1.19}{100}=12.39\text{M}$
$\text{M}_1\text{V}_1=\text{M}_2\text{V}_2$
$\Rightarrow12.39\times\text{V}_1=0.1\times100\text{ml}$
$\Rightarrow\text{V}_1=\frac{0.1\times1000}{12.39}\text{M}$
$\Rightarrow\text{V}_1=8.07\text{ml}.$
View full question & answer→Question 593 Marks
A compound on analysis was found to contain C = 34.6%, H = 3.85%, and 0 = 61.55%. Calculate its empirical formula.
AnswerCalculation of the empirical formula.
| Elemenl |
Percentage |
Atomic mass |
Gram atoms (Moles) |
Molar ratio |
Simplest whole numbar ratio |
| C |
34.6 |
12 |
$\frac{34.6}{12}=2.88$ |
$\frac{2.88}{2.88}=1$ |
3 |
| H |
3.85 |
1 |
$\frac{3.85}{1}=3.85$ |
$\frac{3.85}{2.88}=1.335\text{ or }\frac{4}{3}$ |
4 |
| O |
61.65 |
16 |
$\frac{61.55}{16}=3.85$ |
$\frac{3.85}{2.88}=1.335\text{ or }\frac{4}{3}$ |
4 |
$\therefore$ The simplest whole number ratios of the different elements are: C : H : 0 :: 3 : 4 : 4 and the empirical formula of the compound = $\mathrm{C}_3 \mathrm{H}_4 \mathrm{O}_4$. View full question & answer→Question 603 Marks
If the density of methanol is $0.793 \mathrm{~kg} \mathrm{~L}^{-1}$, what is its volume needed for making 2.5 L of its 0.25 M solution?
AnswerMolar mass of methanol $\left(\mathrm{CH}_3 \mathrm{OH}\right)=(1 \times 12)+(4 \times 1)+(1 \times 16)$
$=32 \mathrm{~g} \mathrm{~mol}^{-1}$
$=0.032 \mathrm{~kg} \mathrm{~mol}^{-1}$
Molarity of methanol solution $=\frac{0.793 \mathrm{~kg} \mathrm{~L}^{-1}}{0.032 \mathrm{~kg} \mathrm{~mol}^{-1}}$
$=24.78 \mathrm{~mol} \mathrm{~L}^{-1}$
(Since density is mass per unit volume)
Applying,
$M_1 V_1=M_2 V_2$
(Given solution) (Solution to be prepared)
$\left(24.78 \mathrm{~mol} \mathrm{~L}^{-1}\right) \mathrm{V}_1=(2.5 \mathrm{~L})\left(0.25 \mathrm{~mol} \mathrm{~L}^{-1}\right)$
$\mathrm{V}_1=0.0252 \mathrm{~L}$
$\mathrm{~V}_1=25.22 \mathrm{~mL}$
View full question & answer→Question 613 Marks
A crystalline salt when heated becomes anhydrous and loses 51.2% of its weight. The anhydrous salt on analysis gave the percentage composition as Mg = 20.0%, S = 26.66% and O = 53.33%.
Answer
| Element |
% of mass |
Atomic mass |
Relative no. of moles element |
Simple molar ratio |
| Mg |
20 |
24 |
$\frac{20}{24}=0.8333$ |
$\frac{0.8333}{0.8333}=1$ |
| S |
26.66 |
32 |
$\frac{26.66}{32}=0.8125$ |
$\frac{0.8125}{0.8333}=1$ |
| O |
53.33 |
16 |
$\frac{53.33}{16}=3.333125$ |
$\frac{3.333}{0.8333}=4$ |
The empirical formula of the anhydrous salt comes out to be $\mathrm{MgSO}_4$. Empirical formula mass $=120$.
Molecular mass $=120$.
Hence, molecular formula $=\mathrm{MgSO}_4$.
As crystalline salt on becoming anhydrous loses $51.2 \%$ by mass, this means 48.8 g of anhydrous salt contains $\mathrm{H}_2 \mathrm{O}=$
51.2 g . Therefore, 120 g of anhydrous salt contains
$=\frac{51.2}{48.8} \times 120 \mathrm{~g}=126 \mathrm{~g} \frac{126}{18}$ molecules $=7 \mathrm{H}_2 \mathrm{O}$ molecules
Hence, molecular formula of crystalline salt $=\mathrm{MgSO}_4 \cdot 7 \mathrm{H}_2 \mathrm{O}$ View full question & answer→Question 623 Marks
Calculate the concentration of nitric acid in moles per litre in a sample which has a density, $1.41 g mL ^{-1}$ and the mass per cent of nitric acid in it being $69 \%$.
AnswerMass percent of nitric acid in the sample = 69% [Given]
Thus, 100g of nitric acid contains 69g of nitric acid by mass.
Molar mass of nitric acid ($HNO_3$).
$= {1 + 14 + 3(16)}g mol^{–1}$
$= 1 + 14 + 48$
$= 63g mol^{–1}$
$\therefore$ Number of moles in 69g of $HNO_3.$
$=\frac{69\text{g}}{63\text{g mol}^{-1}}$
$=1.095\text{ mol}$
Volume of 100g of nitric acid solution,
$=\frac{\text{Mass of solution}}{\text{density of solution}}$
$=\frac{100\text{g}}{1.41\text{g mL}^{-1}}$
$=70.92\text{mL}=70.92\times10^{-3}\text{L}$
Concentration of nitric acid,
$=\frac{1.095\text{mole}}{70.92\times10^{-3}\text{L}}$
$=15.44\text{mol/L}$
$\therefore$ Concentration of nitric acid = 15.44mol/L.
View full question & answer→Question 633 Marks
How much copper can be obtained from 100g of copper sulphate ($\mathrm{CuSO}_4$)?
Answer1 mole of $\mathrm{CuSO}_4$ contains 1 mole of Cu .
Molar mass of $\mathrm{CuSO}_4$
$=(63.5)+(32.00)+4(16.00)$
$=63.5+32.00+64.00$
$=159.5 \text { gram }$
159.5 gram of $\mathrm{CuSO}_4$ contains 63.5 gram of Cu .
Therefore, 100g of $\mathrm{CuSO}_4$ will comtain $\frac{63.5\times100\text{g}}{159.5}$ of Cu.
$=\frac{63.5\times100\text{g}}{159.5}$
$=39.81\text{g}$
View full question & answer→Question 643 Marks
How many grams of $Cl _2$, are required to completely react with 0.4 g of H, to yield HCl ? Also, calculate the amount of HCl formed.
Answer$\text{H}_2\text{(g)} \ \ \ \ \ \ \ \ +\ \ \ \ \ \ \ \text{Cl}_2\text{(g)} \xrightarrow{\ \ \ \ \ \ \ } 2\text{HCl(g)}\$2\times1=2)\ \ \ \ \ (2\times35.5=71)$
Now, 2 g of $H _2$ reacts with 71 g of $Cl _2$ to give 73 g of HCl .
So, 0.4 g of $H _2$ reacts with $\frac{71}{2} \times 0.4=14.2 g$ of $Cl _2$ to given
$\frac{73 \times 0.4}{2}=14.6 g$ of HCl
Therefore, 14.2 g of $Cl _2$ and 14.6 g of HCl are required.
View full question & answer→Question 653 Marks
Hydrogen reacts with nitrogen to produce ammonia according to the equation:
$3H_2(g) + N_2(g) → 2NH_3(g)$
Determine How much ammonia would be produced if 100g of $N_2$ reacts?
Answer$3\text{H}_2\text{(g)}\ \ \ \ +\ \ \ \ \ \text{N}_2\text{(g)}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }2\text{NH}_3\text{(g)}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2\times14=28\text{g})\ \ \ (2\times17=34\text{g})$
Now, 28g of $N_2$, reacts with hydrogen to form 34g of $NH_3$
Hence, 100g of $N_2$ reacts with hydrogen to form
$\frac{34}{28}\times100=\frac{3400}{28}=121.4\text{g}.$
View full question & answer→Question 663 Marks
Calculate:
- Mass in grams of 5.8mol of $N_2O.$
- Number of moles in 8.0g of $O_2.$
- Molar mass of 11.2L at STP weighs 8.5g.
Answer1 mole of $H_2O = 28 + 16 = 44g$
$5.8$ moles of $H_2O = 44 \times 5.8g = 255.2g$
Number of moles $=\frac{\text{Given mass}}{\text{Molar mass of O}_2}$
$=\frac{8.0\text{g}}{32\text{g}}=0.25\text{ mole.}$
11.2L of $NH_2$ at STP weighs $= 8.5g$
$22.4 L$ of $NH_2$ at STP weighs $= 17g$
Molar mass $= 17g mol^{-1}$
[$\because$ 1 mol of every gas at STP has V = 22.4 at STP]
View full question & answer→Question 673 Marks
What is the concentration of sugar $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)$ in $\mathrm{mol}^{-1}$ if its 20 g are dissolved in enough water to make a final volume up to 2L?
AnswerMolarity (M) of a solution is given by,
$=\frac{\text{Number of moles of solute}}{\text{Volume of solution in Litres}}$
$=\frac{\text{Mass of suger/molar mass of sugar}}{2\text{L}}$
$=\frac{20\text{g}/[(12\times12)+(1\times22)+(11\times16)]\text{g}}{2\text{L}}$
$=\frac{20\text{g}/342\text{g}}{2\text{L}}$
$=\frac{0.0585\text{ mol}}{2\text{L}}$
$= 0.02925 \text{ mol L}^{–1}$
$\therefore$ Molar concentration of sugar = 0.02925 mol $\text{L}^{–1}$
View full question & answer→Question 683 Marks
Air contains $20\%$ oxygen by volume. Calculate the theoretical volume of air which will be required for burning completely $500\ m$ of acetylene gas.All volumes are measured under the same conditions of temperature and pressure.
Answer
| $2C_2H_2$ |
$+5O_2$ |
$→4CO_2$ |
$+2H_2O(g)$ |
| 2vol |
5vol |
4vol |
2vol (Gay-Lussac's Law) |
| Or 1vol |
$\frac{5}{2}\text{vol}$ |
$\frac{4}{2}\text{vol}$ |
$\frac{2}{2}\text{vol}$ |
| $500m^3$ |
$\frac{5}{2}\times500\text{cm}^3$ |
$\frac{4}{2}\times500\text{cm}^3$ |
$\frac{2}{2}\times500\text{cm}^3$ |
| $500m^3$ |
$1250m^3$ |
$1000m^3$ |
$500m^3$ |
Thus, $1250m^3$ oxygen is required for burning $500m^3$ of acetylene. But the percentage of oxygen in air is 20%
$\therefore$ Volume of air required $= 1250 \times \frac{100}{20} = 6250\text{m}^3.$ View full question & answer→Question 693 Marks
Calculate the mass percent of different elements present in sodium sulphate ($\mathrm{Na}_2 \mathrm{SO}_4$).
AnswerThe molecular formula of sodium sulphate is $\mathrm{Na}_2 \mathrm{SO}_4$. Molar mass of $\mathrm{Na}_2 \mathrm{SO}_4=[(2 \times 23.0)+(32.066)+4(16.00)]$
$=142.066 \mathrm{~g}$
Mass percent of an element $=\frac{\text{Mass of that element in the compound}}{\text{Molar mass of compoud}}\times100$
$\therefore$ Mass percent of sodium:
$=\frac{46.0\text{g}}{142.066\text{g}}\times100$
$=32.379$
$=32.4\%$
Mass percent of sulphur:
$=\frac{32.066\text{g}}{142.066\text{g}}\times100$
$=22.57$
$=22.6\%$
Mass percent of oxygen:
$=\frac{64.0\text{g}}{142.066\text{g}}\times100$
$=45.049$
$=45.05\%$
View full question & answer→Question 703 Marks
Calculate the atomic mass (average) of chlorine using the following data:
|
|
% Natural Abundance
|
Molar Mass
|
| ${ }^{35} \mathrm{Cl}$ |
75.77
|
34.9689
|
| ${ }^{35} \mathrm{Cl}$ |
24.23
|
36.9659
|
AnswerThe average atomic mass of chlorine
$=\begin{bmatrix}\begin{pmatrix}\text{Fractional abundance}\\\text{ of }\ ^{35}\text{Cl}\end{pmatrix}\begin{pmatrix}\text{Molar mass}\\\text{of }\ ^{35}\text{Cl} \end{pmatrix}\\+\begin{pmatrix}\text{Fractional abundance}\\\text{ of }\ ^{35}\text{Cl}\end{pmatrix}\begin{pmatrix}\text{Molar mass}\\\text{of }\ ^{35}\text{Cl} \end{pmatrix}\end{bmatrix}$
$=\Big[\Big\{\Big(\frac{75.77}{100}\Big)(34.9689)\Big\}+=\Big[\Big\{\Big(\frac{24.23}{100}\Big)(36.9659)\Big\}\Big]$
$= 26.4959 + 8.9568$
$= 35.4527\text{u}$
$\therefore$ The average atomic mass of chlorine = 35.4527u.
View full question & answer→Question 713 Marks
Match Cloumn I with Column II.
|
S. No
|
Cloumn I
|
S. No
|
Cloumn II
|
|
1.
|
88g of $CO_2$
|
(i)
|
0.25mol
|
|
2.
|
$6.022 \times 10^{23}$ molecules of $H_2O$
|
(ii)
|
2mol
|
|
3.
|
5.6 litres of $O_2$ at STP
|
(iii)
|
1mol
|
|
4.
|
96g of $O_2$
|
(iv)
|
$6.022 \times 10^{23}$ molecules
|
|
5.
|
1mol of any gas
|
(v)
|
3mol
|
Answer
|
S. No
|
Cloumn I
|
S. No
|
Cloumn II
|
|
1.
|
88g of $CO_2$
|
(ii)
|
2mol
|
|
2.
|
$6.022 \times 10^{23}$ molecules of $H_2O$
|
(iii)
|
1mol
|
|
3.
|
5.6 litres of $O_2$ at STP
|
(i)
|
0.25mol
|
|
4.
|
96g of $O_2$
|
(v)
|
3mol
|
|
5.
|
1mol of any gas
|
(iv)
|
$6.022 \times 10^{23}$ molecules
|
View full question & answer→Question 723 Marks
A box contains some identical red coloured balls, labelled as A, each weighing 2 grams. Another box contains identical blue coloured balls, labelled as B, each weighing 5 grams. Consider the combinations $AB, AB_{2,} A_2B$ and $A_2B_3$ and show that law of multiple proportions is applicable.
AnswerAccording to the law of multiple pro portion, when two elements comvune to from two or more compoundds, then the different massses one element, which combine with a fixed mass of the other, bear a simple ratio to one another.
1g of A combines with $\frac{5}{2}\text{g}$ of B = 2.5g B
For $AB_2$
1g of A combines with $\frac{10}{2}\text{g}$ of B = 5g of B
For $A_2B$
1g of A combines wuth $\frac{5}{4}\text{g}$ of B = 1.25g of B
For $A_2B_2$
1g of A combines with $\frac{15}{4}\text{g}$ of B = 3.75g of B
Thus, it is proved thed law mukltiple proportions is applicable.
View full question & answer→Question 733 Marks
Calculate the number of moles in the following masses,
- 1.46 metric ton of Al (1 metric ton $= 10^3kg).$
- 7.9 mg of Ca.
Answer
- 1.46 metric ton of Al $= 1.46 \times 10^3 \times 10^3g of Al$
$= 1.46 \times 10^6g$
Atomic mass of Al = 27
Moles of Al $=\frac{\text{mass of Al}}{\text{atomic mass}}=\frac{1.46\times10^{46}}{27}$
$= 5.41 \times 10^4mol$
- 7.9mg of Ca $= 7.9 \times 10^{-3}g of Ca$
Atomic mass of Ca = 40.1
mole of Ca $=\frac{\text{mass of Ca}}{\text{atomic mass}}=\frac{7.9\times106{-3}}{40.1}$
$=1.97\times10^{-4}\text{mol}.$ View full question & answer→Question 743 Marks
For an actual result of an observation to be 5; two students A and B reported their readings as follows:
|
|
Observation number
|
Average
|
|
1
|
2
|
|
Student A
|
4.95
|
4.93
|
4.94
|
|
Student B
|
4.94
|
5.05
|
4.995
|
Which of the students has made a more precise observation? Is his observation accurate too?
AnswerStudent 'A' has made a more precise observation since the variation in the two readings taken by him is not much.
His observation is precise but is not accurate since his readings are not close to the actual reading which is 5. Student B is more accurate as his average reading is close to actual reading i.e. 5.
View full question & answer→Question 753 Marks
Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively.
Note: Given that the molar mass of the oxide is $159.69 \mathrm{~g} \mathrm{~mol}^{-1}$
Answer$\%$ of iron by mass $=69.9 \%$ [Given]
$\%$ of oxygen by mass $=30.1 \%[$ Given]
Atomic mass of iron $=55.85 \mathrm{amu}$.
Atomic mass of oxygen $=16.00 \mathrm{amu}$.
Relative moles of iron in iron oxide $=\%$ mass of iron by mass/Atomic mass of iron $=69.9 / 55.85=1.25$
Relative moles of oxygen in iron oxide $=\%$ mass of oxygen by mass/Atomic mass of oxygen $=30.01 / 16=1.88$
Simplest molar ratio $=1.25 / 1.25: 1.88 / 1.25$
$\Rightarrow 1: 1.5=2: 3$
$\therefore$ The empirical formula of the iron oxide is $\mathrm{Fe}_2 \mathrm{O}_3$.
Mass of $\mathrm{Fe}_2 \mathrm{O}_3=(2 \times 55.85)+(3 \times 16.00)=159.7 \mathrm{~g} \mathrm{~mol}^{-1} \mathrm{n}=$ Molar mass $/$ Empirical formula mass $=159.7 / 159.6=$ 1(approx)
Thus, Molecular formula is same as Empirical Formula i.e. $\mathrm{Fe}_2 \mathrm{O}_3$.
View full question & answer→Question 763 Marks
QUESIION Potassium superoxide, $KO _2$ is used in rebreathing gas masses to generate oxygen
$4 KO_2(s)+2 H_2 O(l) \rightarrow 4 KOH(s)+3 O_2(g)$
If a reaction vessel contains $0.15{mol KO _2}^2$ and $0.10 mol H _2 O$, what is limiting reactant. Bow many moles of oxygen can be produced?
Answer$4$ moles of $KO _2$ reacts with 2 moles of $H _2 O$
$0.15$ mole of $KO _2$ will react with $\frac{2}{4} \times 0.15=0.075$ moles of water.
But we have 0.10 mol of $H _2 O$
$\therefore H _2 O$ is excess reagent, $KO _2$ is limiting reagent.
$4$ moles of $KO _2$ gives 3 moles of $O _2$
$0.15$ mole of $KO _2$ will given $\frac{3}{4} \times 0.15=\frac{0.45}{4}=0.1125$ moles of $O _2$.
View full question & answer→Question 773 Marks
A 0.005 cm thick coating of copper is deposited on a plate of $0.5 \mathrm{~m}^2$ total area. Calculate the number of copper atoms deposited on the plate 0 (density of copper $=7.2 \mathrm{~g} \mathrm{~cm}^{-3}$, atomic mass $=63.5$ ).
AnswerArea of plate $=0.5 \mathrm{~m}^2=0.5 \times 10^4 \mathrm{~cm}^2$
Thickness of coating $=0.005 \mathrm{~cm}$
Volume of copper deposited $=0.5 \times 10^4 \times 0.005=25 \mathrm{~cm}^3$
Mass of copper deposited $=25 \times 7.2=180 \mathrm{~g}$
Now, 63.5 g of copper contains atoms $=6.022 \times 10^{23}$
$\therefore 180 \mathrm{~g}$ of copper will contain atoms $=\frac{6.022 \times 10^{23}}{36.5} \times 180$
$=1.71 \times 10^{23}$ atoms.
View full question & answer→