Solutions and Colligative Properties 1 Question 1
1. The mole fraction of a solvent in aqueous solution of a solute is 0.8 . The molality (in mol $\mathrm{kg}^{-1}$ ) of the aqueous solution is
(2019 Main, 12 April I)
(a) $13.88 \times 10^{-2}$
(b) $13.88 \times 10^{-1}$
(c) 13.88
(d) $13.88 \times 10^{-3}$
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Solution:
- Key Idea Molality $(m)=\frac{\text { Mass of solute }\left(w_{2}\right) \times 1000}{\text { Molar mass of solute }\left(M_{2}\right) \times}$ mass of solvent $\left(w_{1}\right)$
$$ m=\frac{w_{2}}{M_{2}} \times \frac{1000}{w_{1}} $$
and also,
$$ m=n_{2} \times \frac{1000}{n_{1} \times M_{1}} $$
$X_{\text {solvent }}=0.8$ (Given) It means that $n_{\text {solvent }}\left(n_{1}\right)=0.8$ and $n_{\text {solute }}\left(n_{2}\right)=0.2$
Using formula $m=n_{2} \times \frac{1000}{n_{1} \times M_{1}}=0.2 \times \frac{1000}{0.8 \times 18}=13.88 \mathrm{~mol} \mathrm{~kg}^{-1}$
Key Idea Molality is defined as number of moles of solute per $\mathrm{kg}$ of solvent.
$$ m=\frac{w_{2}}{M w_{2}} \times \frac{1000}{w_{1}} $$
$w_{2}=$ mass of solute, $M w_{2}=$ molecular mass of solute $w_{1}=$ mass of solvent.
The molality of $20 %$ (mass/mass) aqueous solution of KI can be calculated by following formula.
$$ m=\frac{w_{2} \times 1000}{M w_{2} \times w_{1}} $$
$20 %$ aqueous solution of $\mathrm{KI}$ means that $20 \mathrm{gm}$ of $\mathrm{KI}$ is present in $80 \mathrm{gm}$ solvent.
$$ m=\frac{20}{166} \times \frac{1000}{80}=1.506 \approx 1.51 \mathrm{~mol} / \mathrm{kg} $$
