Solutions and Colligative Properties - Result Question 1
1. The mole fraction of a solvent in aqueous solution of a solute is $0.8$ . The molality (in $\mathrm{mol}$ $ \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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Answer:
Correct Answer: 1. (c)
Solution:
Key Idea: Molality $(m)=\dfrac{\text { Mass of solute }\left(w_2\right) \times 1000}{\text { Molar mass of solute }\left(M_2\right) \times \text { mass of solvent }\left(w_1\right)} $
$ m=\dfrac{w_2}{M_2} \times \dfrac{1000}{w_1} $
and also, $ \quad m=n_2 \times \dfrac{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 $m=n_2 \times \dfrac{1000}{n_1 \times M_1}=0.2 \times \dfrac{1000}{0.8 \times 18}=13.88 \mathrm{~mol} \mathrm{~kg}^{-1}$
BETA
