Answer

Mass percentage of ${C_6} {H_6}$ $=\dfrac{\text { Mass of } {C_6} {H_6}}{\text { Total mass of the solution }} \times 100 \% $

$$ \begin{aligned} & =\dfrac{\text { Mass of } {C_6} {H_6}}{\text { Mass of } {C_6} {H_6}+\text { Mass of } {CCl_4}} \times 100 \% \\ & =\dfrac{22}{22+122} \times 100 \% \\ & =15.28 \% \end{aligned} $$

Mass percentage of ${CCl_4}$ $ =\dfrac{\text { Mass of } {CCl_4}}{\text { Total mass of the solution }} \times 100 \% $

Mass percentage of ${CCl_4}$ $ =\dfrac{\text { Mass of } {CCl_4}}{\text { Mass of } {C_6} {H_6}+\text { Mass of } {CCl_4}} \times 100 \% $

$ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad =\dfrac{122}{22+122} \times 100 \%$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad=84.72 \%$

Alternatively,

Mass percentage of ${CCl_4}=(100-15.28) \%$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad=84.72 \%$

Answer

Let the total mass of the solution be $100 {~g}$ and the mass of benzene be $30 {~g}$.

$\therefore$ Mass of carbon tetrachloride $=(100-30) {g}$ $=70 {~g}$

Molar mass of benzene $({C_6} {H_6})=(6 \times 12+6 \times 1) {g} {mol}^{-1}$ $=78 {~g} {~mol}^{-1}$

$\therefore$ Number of moles of ${C_6} {H_6}=\dfrac{30}{78} {~mol}$ $=0.3846 {~mol}$

Molar mass of carbon tetrachloride $({CCl_4})=1 \times 12+4 \times 35.5$ $=154 {~g} {~mol}^{-1}$

$\therefore$ Number of moles of ${CCl_4}=\dfrac{70}{154} {~mol}$ $=0.4545 {~mol}$

$ \begin{aligned} &\text {Thus, the mole fraction of }{C_6} {H_6} \text { is given as } =\dfrac{\text { Number of moles of } {C_6} {H_6}}{\text { Number of moles of } {C_6} {H_6}+\text { Number of moles of } {CCl_4}} \end{aligned} $

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad=\dfrac{0.3846}{0.3846+0.4545}$

$ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad= 0.458$

Answer

Molarity is given by:

$$ \text { Molarity }=\dfrac{\text { Moles of solute }}{\text { Volume of solution in litre }} $$

(a) Molar mass of ${Co}({NO_3})_{2} \cdot 6 {H_2} {O}=59+2(14+3 \times 16)+6 \times 18$ $=291 {~g} {~mol}^{-1}$

$\therefore$ Moles of ${Co}({NO_3})_{2} \cdot 6 {H_2} {O}=\dfrac{30}{291} {~mol}$ $=0.103 {~mol}$

Therefore, molarity $=\dfrac{0.103 {~mol}}{4.3 {~L}}$ $=0.023\hspace{0.5mm} {M}$

(b) Number of moles present in $1000 {~mL}$ of $0.5 {M} {H_2} {SO_4}=0.5 {~mol}$

$\therefore$ Number of moles present in $30 {~mL}$ of $0.5 {M} {H_2} {SO_4}=\dfrac{0.5 \times 30}{1000} {~mol}$ $=0.015 {~mol}$

Therefore, molarity $ =\dfrac{0.015}{0.5 {~L}} {~mol} $ $=0.03 {M}$

Answer

Molar mass of urea $({NH_2} {CONH_2})=2(1 \times 14+2 \times 1)+1 \times 12+1 \times 16$ $=60 {~g} {~mol}^{-1}$

0.25 molar aqueous solution of urea means: $1000 {~g}$ of water contains $0.25 {~mol}=(0.25 \times 60) {g}$ of urea

$ \quad\quad \quad\quad \quad\quad \quad\quad \quad\quad \quad\quad\quad\quad \quad\quad \quad\quad \quad\quad \quad\quad \quad\quad \quad\quad \quad\quad \quad\quad= 15 {~g}$ of urea

That is, $(1000+15) {g}$ of solution contains $15 {~g}$ of urea

Therefore, $2.5 {~kg}(2500 {~g})$ of solution contains $ =\dfrac{15 \times 2500}{1000+15} {~g} $

$\quad\quad \quad\quad \quad\quad \quad\quad \quad\quad \quad\quad\quad\quad \quad\quad \quad\quad \quad =36.95 {~g}$

$ \quad\quad \quad\quad \quad\quad \quad\quad \quad\quad \quad\quad\quad\quad \quad\quad \quad\quad \quad = 37 {~g}$ of urea (approximately)

Hence, mass of urea required $=37 {~g}$

Note : There is a slight variation in this answer and the one given in the NCERT textbook.

Answer

(a) Molar mass of ${KI}=39+127=166 {~g} {~mol}^{-1}$

$20 \%$ (mass/mass) aqueous solution of ${KI}$ means $20 {~g}$ of ${KI}$ is present in $100 {~g}$ of solution.

That is,$20 {~g}$ of KI is present in $(100-20) {g}$ of water $=80 {~g}$ of water

Therefore, molality of the solution $ =\dfrac{\text { Moles of KI }}{\text { Mass of water in } {kg}} $

$ \quad\quad \quad\quad \quad\quad \quad\quad \quad \quad \quad \qquad = \dfrac{\dfrac{20}{166}}{0.08} {~m}$ $=1.506 {~m}$

$\quad\quad \quad\quad \quad\quad \quad \quad \quad \qquad=1.51 {~m}$ (approximately)

(b) It is given that the density of the solution $=1.202 {~g} {~mL}^{-1}$

$$ Volume=\dfrac{\text { Mass }}{\text { Density }} $$

$$ \quad\quad \quad\qquad=\dfrac{100 {~g}}{1.202 {~g} {~mL}^{-1}}$$

$$\quad\quad \quad\qquad=83.19 {~mL}$$

$$\quad\quad \quad\qquad=83.19 \times 10^{-3} {~L}$$

Therefore, molarity of the solution $ =\dfrac{\dfrac{20}{166} {~mol}}{83.19 \times 10^{-3} {~L}} $

$\quad\quad \quad\qquad\quad\quad \qquad\quad \quad\qquad=1.45\hspace{0.5mm} {M}$

(c) Moles of KI $ =\dfrac{20}{166}=0.12 {~mol} $

Moles of water $ =\dfrac{80}{18}=4.44 {~mol} $

Therefore, mole fraction of ${KI}$ $ =\dfrac{\text { Moles of KI }}{\text { Moles of KI }+ \text { Moles of water }} $ $ =\dfrac{0.12}{0.12+4.44} =0.0263$