Question 15 Marks
A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0L (measured at STP) of this welding gas is found to weigh 11.6g. Calculate.
- Empirical formula.
- Molar mass of the gas.
- Molecular formula.
AnswerAmount of carbon in 3.38g $\text{CO}_2=\frac{12}{44}\times3.38\text{g}=0.9218\text{g}$ Amount of hydrogen in 0.690g $\text{H}_2\text{O}=\frac{2}{18}\times0.690\text{g}=0.0767\text{g}$ As compound only C and H, Therefore, total mass of the compound, = 0.9218 +0.0767g = 0.9985g % of C in the compound $=\frac{0.9218}{0.9985}\times100=92.32$ % of H in the compound $=\frac{0.0767}{0.9985}\times100=7.68$ Calculation of Empirical Formula.
| Element |
% by mass |
Atomic mass |
Moles of the element |
Simplest molar ratio |
Simplest whole no. molar ratio |
| C |
92.32 |
12 |
$\frac{92.32}{12}=7.69$ |
1 |
1 |
| H |
7.68 |
1 |
$\frac{7.68}{1}=7.68$ |
1 |
1 |
$\therefore$ Empirical formula = CH 10.0L of the gas at STP weight = 11.6g
$\therefore$ 22.4L of the gas at S.T.P. will weight $=\frac{11.6}{10.0}\times22.4=25.984\text{g}\approx26\text{g}$
$\therefore$ Moral mass $= 26g\ mol^{-1} $
Emprirical formula mass of CH = 12 + 1 = 13
$\therefore\text{n}=\frac{\text{Molecular mass}}{\text{E.F. mass}}$
$\therefore\text{Molecular formula }=2\times\text{CH}=\text{C}_2\text{H}_2$ View full question & answer→Question 25 Marks
Match the following prefixes with their multiples:
| |
Prefix
|
Multiples
|
|
(i)
|
micro
|
$10^6$
|
|
(ii)
|
deca
|
$10^9$
|
|
(iii)
|
mega
|
$10^{-6}$
|
|
(iv)
|
giga
|
$10^{-15}$
|
|
(v)
|
femto
|
$10$
|
Answer
|
|
Prefix
|
Multiples
|
|
(i)
|
micro
|
$10^{–6}$
|
|
(ii)
|
deca
|
10
|
|
(iii)
|
mega
|
$10^6$
|
|
(iv)
|
giga
|
$10^9$
|
|
(v)
|
femto
|
$10^{–15}$
|
View full question & answer→Question 35 Marks
Determine the empirical formula of an oxide of iron, which has $69.9\%$ iron and $30.1\%$ dioxygen by mass.
AnswerPercent of Fe by mass = 69.9% [As given above]
Percent of $O_2$ by mass = 30.1% [As given above]
Relative moles of Fe in iron oxide:
$=\frac{\text{percent of iron by mass}}{\text{Atomic mass of iron}}$
$=\frac{69.9}{55.85}$
$=1.25$
Relative moles of O in iron oxide:
$=\frac{\text{percent of oxygen by mass}}{\text{Atomic mass of oxygen}}$
$=\frac{30.1}{16.00}$
$=1.88$
Simplest molar ratio of Fe to O:
= 1.25 : 1.88
= 1 : 1.5
= 2 : 3
Therefore, empirical formula of iron oxide is $Fe_2O_3$.
View full question & answer→Question 45 Marks
Calcium carbonate reacts with aqueous HCl to give $\mathrm{CaCl}_2$ and $\mathrm{CO}_2$ according to the reaction, $\mathrm{CaCO}_{3(\mathrm{~s})}+2 \mathrm{HCl}_{(\mathrm{aq})} \rightarrow$ $\mathrm{CaCl}_{2(\mathrm{aq})}+\mathrm{CO}_{2(g)}+\mathrm{H}_2 \mathrm{O}_{(l)}$
What mass of $\mathrm{CaCO}_3$ is required to react completely with 25 mL of 0.75 M HCl ?
Answer0.75 M of $\mathrm{HCl} \equiv 0.75 \mathrm{~mol}$ of HCl are present in 1 L of water
$\equiv\left[(0.75 \mathrm{~mol}) \times\left(36.5 \mathrm{~g} \mathrm{~mol}^{-1}\right)\right] \mathrm{HCl}$ is present in 1 L of water
$\equiv 27.375 \mathrm{~g}$ of HCl is present in 1 L of water
Thus, 1000 mL of solution contains 27.375 g of HCl .
$\therefore$ Amount of HCl present in 25 mL of solution
$=\frac{27.375 \mathrm{~g}}{1000 \mathrm{~mL}} \times 25 \mathrm{~mL}$
$=0.6844 \mathrm{~g}$
From the given chemical equation,
$\mathrm{CaCO}_{3(\mathrm{~s})}+2 \mathrm{HCl}_{(\mathrm{aq})}-\mathrm{CaCl}_{2(\mathrm{aq})}+\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}$
2 mol of $\mathrm{HCl}(2 \times 36.5=73 \mathrm{~g})$ react with 1 mol of $\mathrm{CaCO}_3(100 \mathrm{~g})$.
$\therefore$ Amount of $\mathrm{CaCO}_3$ that will react with 0.6844 g .
$=100 / 73 \times 0.684410073 \times 0.6844$
$=0.9375 \mathrm{~g}$
View full question & answer→Question 55 Marks
Express the following in the scientific notation:
-
0.0048
-
234, 000
-
8008
-
500.0
-
6.0012
Answeri. $0.0048=4.8 \times 10^{-3}$
ii. $234,000=2.34 \times 10^5$
iii. $8008=8.008 \times 10^3$
iv. $500.0=5.000 \times 10^2$
v. $6.0012=6.0012 \times 10^0$
View full question & answer→Question 65 Marks
Chlorine is prepared in the laboratory by treating manganese dioxide $\left(\mathrm{MnO}_2\right)$ with aqueous hydrochloric acid according to the reaction
$4 \mathrm{HCl}_{(\mathrm{aq})}+\mathrm{MnO}_{2(\mathrm{~s})} \rightarrow 2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}+\mathrm{MnCl}_{2(\mathrm{aq})}+\mathrm{Cl}_{2(\mathrm{~g})}$
How many grams of HCl react with 5.0 g of manganese dioxide?
Answer$1 \mathrm{~mol}[55+2 \times 16=87 \mathrm{~g}] \mathrm{MnO}_2$ reacts completely with $4 \mathrm{~mol}[4 \times 36.5=146 \mathrm{~g}]$ of HCl .
$\therefore 5.0 \mathrm{~g}$ of $\mathrm{MnO}_2$ will react with
$=\frac{146 \mathrm{~g}}{87 \mathrm{~g}} \times 5.0 \mathrm{~g} \text { of } \mathrm{HCl}$
$=8.4 \mathrm{~g} \text { of } \mathrm{HCl}$
Hence, 8.4 g of HCl will react completely with 5.0 g of manganese dioxide.
View full question & answer→Question 75 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 85 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 95 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?
Answer$\text { Molar mass of methanol }\left(\mathrm{CH}_3 \mathrm{OH}\right)=(1 \times 12)+(4 \times 1)+(1 \times 16)$
$=32 \mathrm{~g} \mathrm{~mol}^{-1}$
$=0.032 \mathrm{~kg} \mathrm{~mol}^{-1}$
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,
$\mathrm{M}_1 \mathrm{~V}_1=\mathrm{M}_2 \mathrm{~V}_2$
(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 105 Marks
Calculate the concentration of nitric acid in moles per litre in a sample which has a density, $1.41 \mathrm{~g} \mathrm{~mL}^{-1}$ and the mass per cent of nitric acid in it being $69 \%$.
AnswerMass percent of nitric acid in the sample $=69 \%$ [Given]
Thus, 100 g of nitric acid contains 69 g of nitric acid by mass.
Molar mass of nitric acid $\left(\mathrm{HNO}_3\right)$.
$=\{1+14+3(16)\} \mathrm{g} \mathrm{~mol}^{-1}$
$=1+14+48$
$=63 \mathrm{~g} \mathrm{~mol}^{-1}$
$\therefore$ Number of moles in 69 g of $\mathrm{HNO}_3$.
$=\frac{69\text{g}}{63\text{g mol}^{-1}}$
$=1.095\text{ mol}$
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 115 Marks
How much copper can be obtained from 100 g of copper sulphate $\left(\mathrm{CuSO}_4\right) ?$
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, 100 g of $\mathrm{CuSO}_4$ will comtain $\frac{63.5 \times 100 \mathrm{~g}}{159.5}$ of Cu .
$=\frac{63.5\times100\text{g}}{159.5}$
$=39.81\text{g}$
View full question & answer→Question 125 Marks
What is the concentration of sugar $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)$ in $\mathrm{mol} \mathrm{~L}^{-1}$ 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 $=0.02925 \mathrm{~mol} \mathrm{~L}^{-1}$
View full question & answer→Question 135 Marks
Calculate the mass percent of different elements present in sodium sulphate $\left(\mathrm{Na}_2 \mathrm{SO}_4\right)$.
AnswerThe molecular formula of sodium sulphate is $\mathrm{Na}_2 \mathrm{SO}_4$.
$\text { Molar mass of } \mathrm{Na}_2 \mathrm{SO}_4=[(2 \times 23.0)+(32.066)+4(16.00)]$
$=142.066 \mathrm{~g}$
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 145 Marks
Calculate the atomic mass (average) of chlorine using the following data:
|
|
% Natural Abundance
|
Molar Mass
|
| ${ }^{35} \mathrm{Cl}$ |
75.77
|
34.9689
|
| ${ }^{37} \mathrm{Cl}$ |
24.23
|
36.9659
|
AnswerThe average atomic mass of chlorine
$=\begin{bmatrix}\begin{pmatrix}\text{Fractional abundance}\\\text{ of }\ ^{35}\text{C}1\end{pmatrix}\begin{pmatrix}\text{Molar mass}\\\text{of }\ ^{35}\text{C}1 \end{pmatrix}\\+\begin{pmatrix}\text{Fractional abundance}\\\text{ of }\ ^{35}\text{C}1\end{pmatrix}\begin{pmatrix}\text{Molar mass}\\\text{of }\ ^{35}\text{C}1 \end{pmatrix}\end{bmatrix}$
$=\Big[\Big\{\Big(\frac{75.77}{100}\Big)(34.9689)\Big\}+=\Big[\Big\{\Big(\frac{24.23}{100}\Big)(36.9659)\Big\}\Big]$
$= 26.4959 + 8.9568$
$= 35.4527\text{u}$
$\therefore$ The average atomic mass of chlorine = 35.4527u.
View full question & answer→Question 155 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 $159.69g\ mol^{-1}$
Answer% of iron by mass = 69.9% [Given]
% of oxygen by mass = 30.1% [Given]
Atomic mass of iron = 55.85amu.
Atomic mass of oxygen = 16.00amu.
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 $Fe_2O_3$.
Mass of $Fe_2O_3 = (2 \times 55.85) + (3 \times 16.00) = 159.7g mol^{-1}n$ = Molar mass/Empirical formula mass = $159.7/159.6 = 1(approx)$
Thus, Molecular formula is same as Empirical Formula i.e. $Fe_2O_3$.
View full question & answer→Question 165 Marks
The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction 2A + 4B → 3C + 4D, when 5 moles of A react with 6 moles of B, then
- Which is the limiting reagent?
- Calculate the amount of C formed?
AnswerThe reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction 2A + 4B → 3C + 4D, when 5 moles of A react with 6 moles of B, then
The given reaction is:
$2\text{A}+4\text{B}\rightarrow3\text{C}+4\text{D}\\^{2\text{mol}}\ \ \ ^{4\text{mol}}\ \ \ \ \ ^{3\text{mol}}\ \ \ \ ^{4\text{mol}}$
Given moles of A and B are 5 and 6 moles reapectively.
Case I:
Let reactant (A) is completely consumed.
2mol A gives 3mol C
$\therefore$ 5mol A will give $\frac{3}{3}\times5\text{mol}$ C = 7.5mol C
Case II:
Let reactant (B) is completely consumed.
4 mol B gives 3mol C
$\therefore$ 6 mol B will give $\frac{3}{4}\times6\text{mol}$ C = 4.5mol C
Since (B) on complete consumption guves lesser amount of product (C),
Hence (B) will be limiting reagent and the amount of (C) formed will be 4.5mol
View full question & answer→Question 175 Marks
A compound on analysis found to contain following percentage composition:
$Na = 43.4\%, C = 11.4\%$ and $O = 45.3\%$. Determine the empirical and molecular formulae.
Given: the relative molecular mass of the compound is $106$.
AnswerGiven results are tabulated as follows:
| Element |
Percentage composition |
Atomic mass |
Relative number of moles |
Simplest molar ratio |
Simplest whole number molar ratio |
| Na |
43.4 |
23 |
$\frac{43.4}{23}=1.89$ |
$\frac{1.89}{0.94}=3$ |
2 |
| C |
11.3 |
12 |
$\frac{11.3}{12}=0.94$ |
$\frac{0.94}{94}=1$ |
1 |
| O |
45.3 |
16 |
$\frac{45.3}{16}=2.83$ |
$\frac{2.83}{0.94}=3$ |
3 |
The empirical formula of the given compound is $Na_2CO_3$
Now, Empirical formula mass of the compound = (2 × 23u) + (1 × 12u) + (3 × 16u) = 46u + 12u +48u = 106u
Then, $\text{n}=\frac{\text{Molecular mass}}{\text{Empirical formula mass}}$
$=\frac{106\text{u}}{106\text{u}}=1$
Therefore, molecular formula of the given compound is given by,
Molecular formula = 1 × Empirical formula = 1 × $Na_2CO_3 = Na_2CO_3$ View full question & answer→Question 185 Marks
Perform the following calculations and express the results to proper number of significant figures.
- $144.3 m^2 + (2.54 m \times 8.4 m)$.
- $(4.05 \times 10^2mL) - (0.0225 \times 10^2mL)$.
- $(3.50 \times 10^2cm) (4.00 \times 10^6cm)$.
Answer
- $(144.3m^2) + (2.54m \times 8.4m)$
$2.54m \times 8.4m = 21.336$ or $21m^3$
(upto 2 significant figures)
$\ \ \ 144.3\ \text{m}^2\\+\ \ \ \ 21\ \text{m}^2\\\overline{\ \ \ 165.3\ \text{m}^2}$
(As 144.3 contains (1) one digit (SF) after the decimal point.)
- $(4.05 \times 10^2mL) - (0.0225 \times 10^2mL)$
$= 4.0275 \times 10^2mL$ or $4.03 \times 10^2mL$
(upto two decimal place as in 4.05)
- $(3.50 \times 10^2cm) (4.00 \times 10^6cm)$
$= 14.0 \times 10^8 cm^2$ (upto 3 significant figures). View full question & answer→Question 195 Marks
- A sample of salt has the following percentage composition
Fe = 36.76%, S = 21.11 % and O = 42.14%
Calculate the empirical formula of the compound
- What happens if the compound is heated? Write the balanced chemical equation.
Answer
-
| Element |
Percentage |
Atomic mass |
Relative number of moles |
Divide by least |
Simplest ratio |
| Fe |
36.76 |
56 |
$\frac{36.76}{56}=0.65$ |
$\frac{0.65}{0.64}=1$ |
1 |
| S |
21.11 |
32 |
$\frac{21.12}{32}=0.64$ |
$\frac{0.64}{0.64}=1$ |
1 |
| O |
42.11 |
16 |
$\frac{42.11}{16}=2.63$ |
$\frac{2.63}{0.64}=4$ |
4 |
- Empirical formula = $FeSO_4$
$2\text{FeSO}_4\xrightarrow{\text{heat}}\text{Fe}_2\text{O}_3+\text{SO}_2+\text{SO}_3$
Ferrous sulphate, on heating gives $Fe_2O_3, SO_2$ and $SO_3$ gases. View full question & answer→Question 205 Marks
A compound made up of two elements A and B has A = 70%, B = 30%. Their relative number of moles in the compound are $1.25$ and $1.88$. Calculate.
- Atomic masses of the elements A and B.
- Molecular formula of the compound, if its molecular mass is found to be 160.
AnswerRelative number of moles of an element $=\frac{\text{% of the element}}{\text{atomic mass}}$ or atomic mass $=\frac{\%\text{of element}}{\text{relative number of moles}}$ $\therefore$ Atomic mass of $\text{A}=\frac{70}{1.25}=56$ and Atomic mjass of $\text{B}=\frac{30}{1.88}=16$ Calculation of Empirical Formula Relative
| Element |
Relative number of moles |
Simplest molar ratio |
Simplest whole number molar ratio |
| A |
1.25 |
$\frac{1.25}{1.25}=1$ |
2 |
| B |
1.88 |
$\frac{1.88}{1.25}=1.5$ |
3 |
$\therefore$ Empirical formula = $A_2B_3$
Calculation of molecular formula
Empirical formula mass = 2 × 56 + 3 × 16 = 160
$\text{n}=\frac{\text{molecular mass}}{\text{empirical formula mass}}$ $=\frac{160}{160}=1$$\therefore$ Molecular formula = $A_2B_3$ View full question & answer→Question 215 Marks
- $1.84\ g$ of mixture of $CaCO_3$ and $MgCO_3\rightarrow$ is strongly heated till no further loss of mass takes place. The residue weighs 0.96g. Calculate the percentage composition of the mixture.
- What will be the molality of the solution containing $18.25\ g$ of HCl gas in $500\ g$ of water?
Answer
- $\text{CaCO}_3\xrightarrow{\ \ \ \ \ }\text{CaO}+\text{CO}\\100\text{g}\ \ \ \ \ \ \ \ \ \ \ \ \ \ 56\text{g}$
$MgCO_3 MgO+CO_2$
$24-12+48-84 g 24+16=40 g$
Let the mass of CaCO , be ' x ', $MgCO _3$ will be $1- x$. 100 g of $CaCO _3$ will gives 56 g of CaO x g of $CaCO _3$ will gives $\frac{56 x }{100} g$ of CaO
84 g of $MgCO _3$ gives $40\ g$ of MgO
$1- xg$ of $MgCO _3$ gives
$=\frac{40(1-x)}{84} g \text { of } MgO \frac{56 x}{100}+\frac{40(1-x)}{84}=0.96 g$
$84 \times 0.56 x+40 \times 1.84-40 x=0.96 \times 84 \times 100$
$47.04 x-40 x=80.64-73.60$
$7.04 x=7.04$
$\Rightarrow x=1$
$\% \text { of } CaCO_3=100=54.35 \%$
$\% \text { of } MgCO_1=100-54.35-45.65 \%$
- $\text{M}=\frac{\text{Number of moles of solute}}{\text{Litres of solution}}$
$=\frac{\text{Mass of solute}}{\text{Molar mass of solute}}\times\frac{1000}{\text{Mass of solvent in grams}}$
$=\frac{18.25}{36.50}\times\frac{1000}{500}=1\text{mol /kg}=1\text{m}$
$W_B$ is mass of solute, $M_B$ is molar mass of solute,
$W_A$ is mass of solvent in grams. View full question & answer→Question 225 Marks
Sulphuric acid reacts with sodium hydroxide as follows:
$H_2SO_4 + 2NaOH Na_2SO_4 + 2H_2O$
When $1 L$ of $0.1 M$ sulphuric acid solution is allowed to react with $1 L$ of $0.1 M$ sodium hydroxide solution. Calculate the amount of sodium sulphate formed and its molarity in the solution obtained.
AnswerNo. of moles of $\mathrm{H}_2 \mathrm{SO}_4=1 \times 0.1 \mathrm{M}=0.1$ moles
No. of moles of $\mathrm{NaOH}=1 \times 0.1 \mathrm{M}=0.1$ moles
1 mole of $\mathrm{H}_2 \mathrm{SO}_4$ reacts with 2 moles of NaOH
0.1 mole of $\mathrm{H}_2 \mathrm{SO}_4$ reacts with 0.2 moles of NaOH
But we have only 0.1 mole of NaOH
M last of $\mathrm{Na}_2 \mathrm{SO}_4=2 \times 23+32+64=46+32+64$
$=142 \mathrm{~g} \mathrm{~mol}$
$\therefore \mathrm{NaOH}$ is limiting reactant.
2 -moles of NaOH gives 142 g of $\mathrm{Na}_2 \mathrm{SO}_4$
0.1 moles of NaOH gives 142 g of $\mathrm{Na}_2 \mathrm{SO}_4$
$=\frac{142}{2}\times0.1=7.1\text{g}$
$\text{M}=\frac{\text{W}_{\text{B}}}{\text{M}_{\text{B}}}\times\frac{1}{\text{Vol. of solution in L}}$
$=\frac{7.1}{142}\times\frac{1}{2}=0.025\text{mol L}^{-1}$
View full question & answer→Question 235 Marks
Arrange the following in order of their increasing masses in gram (i) One atom of silver, (ii) one gram-atom of nitrogen, (iii) one mole of calcium, (iv) one mole of oxygen molecules, (v) $1023$ atoms of carbon and (vi) one gram of iron.
Answer
- 1 mole of Ag atom = $108g = 6.022 \times 10^{23}$ atoms.
Mass of $6.022 \times 10^{23}$ atoms of Ag = 108g.
Mass of 1 atom of $\text{Ag}=\Big(\frac{108}{6.022\times10^{23}}\Big)$
$=1.1793\times10^{-22}\text{g}$
- Mass of gram atom of N = atomic mass of N in gram = 14.0g.
- Mass of a mole of Ca = atomic mass of Ca in gram = 40.0g.
- Mass of mole of oxygen molecules = molar mass of oxygen in gram = 32.0g.
- Mass of mole of C-atom = $12g = 6.023 \times 10^{23}$ atoms.
Mass of $6.023 \times 10^{23}$ atoms of C = 12g.
Mass of 1 atom of $\text{C}=\Big(\frac{12}{6.023\times10^{23}}\Big)\text{g}$
Mass of $10^{23}$ atoms of $\text{C}=\Big(\frac{12}{6.023\times10^{23}}\Big)$
$=10^{23}=1.992\text{g}$
- Mass of iron = 1.0g. Hence, the required order of m of silver < one gram of iron $<10^{23}$ atoms of C < one-gram atom of nitrogen < one mole of oxygen < one mole of calcium.
View full question & answer→Question 245 Marks
- Commercially available concentrated hydrochloric acid contains $45\%$ HCl by mass.
- What is the molarity of this solution? The density is $1.19g$ mL.
- What volume of cone. HCl is required to make $1.00L$ of $0.24M$ HCl?
- Write the balanced chemical equations for the following:
- $KMnO_4 + NH3 \rightarrow MnO_2 +KOH+ H_2O + N_2$
- $HNO_3 + P_4\rightarrow NO_2 + H_3PO_4 + H_2O$
Answer
-
- 45% solution, means 45g of HCl in 100g of solution.
Then, Mass of the solution = 100g
Volume of the solution $=\frac{100\text{g}}{\text{Density}}=\frac{100\text{g}}{1.19\text{g/ mL}}$
$=840\text{mL}=\frac{84.0}{1000}\text{L}$
Molar mass of HCl = $36.5g mol^{-1}$
No. of moles of HCl dissolved $=\frac{45}{36.5}\text{mol}=1.23\text{mol}$
$\therefore$ Molarity of HCl solution $=\frac{1.23}{\Big(\frac{84}{1000}\Big)\text{L}}=14.64\text{mol L}^{-1}$
- Molarity or conc. HCl sample = 14.64mol/ L
Molarity of HCl solution to be prepared = 0.24mol/ L
Volume of HCl solution to be prepared = 1.00L = 1000mL
Then, using the molarity equation, MV - MV,
$\text{V}_{\text{HCl}}=\frac{0.24}{14.64}=16.40$
Thus, to obtain 1.0L of 0.24M HCl, one should dissolve 16.40mL of conc. HCl to make up the volume to 1.0L.
-
- $2KMnO_4 + 2NH_3 \rightarrow 2MnO_4 + 2KOH + 2H_2O + N_2$
- $2OHNO_3 + P_4\rightarrow 2NO_2 + 4H_3PO_4 + 4H_2O$
View full question & answer→Question 255 Marks
Calculate the moles of NaOH required to neutralize the solution produced by dissolvingg 1.1g $P_4O_6$ in water. Use the following reactions:
$P_4O_6 + 6H_2O \rightarrow 4H_3PO_3$
$2NaOH + H_3PO3 \rightarrow Na_2HPO_3 + 2H_2O$
($Atomic\ mass/ g\ mol^{-1}; P = 31, O = 16$)
Answer$\mathrm{P}_4 \mathrm{O}_6+6 \mathrm{H}_2 \mathrm{O} \rightarrow 4 \mathrm{H}_3 \mathrm{PO}_3$
$2 \mathrm{NaOH}+\mathrm{H}_3 \mathrm{PO}_3 \rightarrow \mathrm{Na}_2 \mathrm{HPO}_3+2 \mathrm{H}_2 \mathrm{O}$
Molecular weight of $\mathrm{P}_4 \mathrm{O}_6=4 \times 31+6 \times 16$
$=124+96=220 \mathrm{~g} \mathrm{~mol}^{-1}$
Number of moles of $\mathrm{P}_4 \mathrm{O}_6=\frac{1.1 \mathrm{~g}}{220}=\frac{1}{200}$ moles $\mathrm{P}_4 \mathrm{O}_6$
$\therefore 4$ moles of $\mathrm{H}_3 \mathrm{PO}_3$ is produced by $\frac{1}{200}=\frac{1}{50}$ moles $\mathrm{P}_4 \mathrm{O}_6$
$\because 1$ moles of $\mathrm{H}_3 \mathrm{PO}_3$ will require $4 \times \frac{1}{200}=\frac{1}{50}$ moles $\mathrm{P}_4 \mathrm{O}_6$ $\frac{1}{200}$ moles $\mathrm{P}_4 \mathrm{O}_6$ Produce $4 \times \frac{1}{200}$ moles $\mathrm{H}_3 \mathrm{PO}_3$
Also, 1 mole of $\mathrm{H}_3 \mathrm{PO}_3$ requires 2 moles of NaOH .
$\frac{1}{50} \text { mole of } \mathrm{H}_3 \mathrm{PO}_3 \text { requires } 2 \times \frac{1}{50}=\frac{1}{25}$
$=0.04 \mathrm{NaOH} .$
View full question & answer→Question 265 Marks
What volume of 0.1 M NaOH solution is required to neutralise 100 ml of concentrated aqueous sulphuric acid which contains $98 \% \mathrm{~H}_2 \mathrm{SO}_4$ by mass. The density of concentrated sulphuric acid solution is $1.84 \mathrm{g} \mathrm{ml}^{-1} \mathrm{NaOH}$ reacts with $\mathrm{H}_2 \mathrm{SO}_4$ according to the following reaction:
$2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{SO} 4 \rightarrow \mathrm{Na}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O}$
(Atomic mass $/ \mathrm{g} \mathrm{mol}^{-1} \mathrm{H}=1, \mathrm{~S}=32, \mathrm{O}=16$ ).
AnswerMolarity $=\frac{\text{Number of moles of solute}}{\text{Litre of solution}}$
$=\frac{\text{Mass of solute}\big(\text{W}_{\text{B}}\big)}{\text{Molar mass of solute}\big(\text{M}_{\text{B}}\big)}\times\frac{1000}{\text{Volume of solution in mL}}$
$=\frac{\text{Mass of solute}\big(\text{W}_{\text{B}}\big)}{\text{Molar mass of solute}\big(\text{M}_{\text{B}}\big)}\times\frac{1000}{\frac{\text{Volume of solution}}{\text{Density of solution}}}$
$=\frac{98}{98}\times\frac{1000}{\frac{\text{Volume of solution}}{\text{Density of solution}}}$
$=\frac{98}{98}\times\frac{1000}{\frac{100}{1.84}}=18.4\text{M}$
$=\frac{98}{98}\times\frac{1000}{100}\times1.84$
$2(\text{M}_1\text{V}_1)_{\text{H}_2\text{SO}_4}=(\text{M}_2\text{V}_2)_{\text{NaOH}}$
$\Rightarrow2\times18.4\text{M}\times100\text{ml}=0.1\text{M}\times\text{V}_2$
$\Rightarrow\text{V}_2=\frac{2\times18.4\times100}{0.1}$
$=36.80\times10^3\text{ml}=36.8\text{L}.$
View full question & answer→Question 275 Marks
Define the law of multiple proportions. Explain it with two examples. How does this law point to the existance of atoms?
AnswerLaw of multiple proportions: When two elements combine to form two or more chemical compounds, then the masses of one of the elements which combine with a fixed mass of the other, bear a simple ratio to one another, e.g., carbon combines with oxygen to form two compounds, namely, carbon dioxide and carbon monoxide. The masses of oxygen which combine with a fixed mass of carbon in $\mathrm{CO}_2$ and CO are 32 and 16 respectively. These masses of oxygen bear a simple ratio of $32: 16$ or $2: 1$ to each other. For example, sulphur combines with oxygen to form two compounds, namely, sulphur trioxide and sulphur dioxide.
The masses of oxygen which combine with a fixed mass of sulphur in $\mathrm{SO}_3$ and $\mathrm{SO}_2$ are 48 and 32 respectively. These masses of oxygen bear a simple ratio of $48: 32$ or $3: 2$ to each other. This law shows that there are constituents whom combine in a definite proportion. These constituents may be atoms. Thus, the law of multiple proportions shows the existence of atoms which combine to form molecules.
View full question & answer→Question 285 Marks
- The density of the water at room temperature is $0.1g/ mL$. How many molecules are there in a drop of water if its volume is $0.05mL$?
- An alloy of iron $(53.6\%)$, nickel $(45.8\%)$ and manganese $(0.6\%)$ has a density of $8.17g cm^{-3}$
Calculate the number of Ni atoms present in the alloy of dimensions $10.0cm \times 20.0cm \times 15.0cm$Answer
- Volume of a drop of water = 0.05mL
Mass of a drop of water = volume × density
= (0.05mL) (1.0g/ mL) = 0.05g
Gram molecular mass of water
(H_2O)= 2 \times 1+ 16 = 18g; 18g of water = 1mol
$\therefore0.05\text{g of water}=\frac{1\text{mol}}{(18\text{g})}\times(0.05\text{g})$
$=0.0028\text{mol}.$
$\because$ 1 mole of water contains molecules = $6.022 \times 10^{23}$
0.0028 mole of water will contain molecules
$= 6.022 \times 10^{23} \times 0.0028 = 1.68 \times 10^{21}$ molecules
- Volume of the alloy
$= (10.0cm) \times (20.0cm) \times (15.0cm) = 3000cm^3$
Mass of the alloy = density × volume
$= (8.17g cm^{-3}) \times (3000cm^3) = 24510g$
Mass of Ni in the alloy $=(24510\text{g})\times\frac{45.8}{100}=11225.6\text{g}$
$\because$ 59g Ni have atoms = $6.022 \times 10^{23}$
Gram atomic mass of Ni = 59g
11225.6g of Ni have atoms
$=6.022\times10^{23}\times\frac{(11225.6\text{g})}{(59.0\text{g})}$
$=1.15\times10^{26}\text{ atoms.}$ View full question & answer→Question 295 Marks
Calculate the mass percent of calcium, phosphorus and oxygen in calcium phosphate $\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2$.
AnswerMass percent of an element $=\frac{\text{atomic mass of the element present in the compound}}{\text{molar mass of the compound}}\times100$
Mass percent of calcium $=\frac{3\times\text{(atomic mass of calcium)}}{\text{molecular mass of Ca}_3(\text{PO}_4)_2}\times100$
$=\frac{120\text{u}}{310\text{u}}\times100=38.7\%$
Mass percent of phosphorus $=\frac{2\times\text{(atomic mass of phosphorus)}}{\text{molecular mass of Ca}_3(\text{PO}_4)_2}\times100$
$=\frac{2\times31\text{u}}{310\text{u}}\times100=20\%$
Mass percent of oxygen $=\frac{8\times\text{(atomic mass of oxygen)}}{\text{molecular mass of Ca}_3(\text{PO}_4)_2}\times100$
$=\frac{8\times16\text{u}}{310\text{u}}\times100=41.29\%$
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