Solutions and Colligative Properties - Result Question 30
####1. The mole fraction of a solvent in aqueous solution of a solute is 0.8 . The molality (in mol $kg^{-1}$ ) 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 $m=n _2 \times \frac{1000}{n _1 \times M _1}=0.2 \times \frac{1000}{0.8 \times 18}=13.88 mol kg^{-1}$
Key Idea Molality is defined as number of moles of solute per $kg$ 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 $KI$ means that $20 gm$ of $KI$ is present in $80 gm$ solvent.
$$ m=\frac{20}{166} \times \frac{1000}{80}=1.506 \approx 1.51 mol / kg $$
