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→