Solutions and Colligative Properties - Result Question 62
33. A solution of a non-volatile solute in water freezes at $-0.30^{\circ} C$. The vapour pressure of pure water at $298 K$ is $23.51 $ $mm $ $Hg$ and $K _f$ for water is $1.86$ $ K$ $ kg$ $ mol^{-1}$. Calculate the vapour pressure of this solution at $298 K$.
(1998,4 M)
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Answer:
Correct Answer: 33. $ (23.44$ $ mm)$
Solution:
$-\Delta T _f=K _f \cdot m _2$
$ \Rightarrow m _2=\dfrac{0.3}{1.86}=0.1613 $
Also, $ m _2=\dfrac{n _2}{n _1} \times \dfrac{1000}{M _1}=0.1613 $
$\Rightarrow \dfrac{n _2}{n _1} =\dfrac{0.1613 \times 18}{1000}=2.9 \times 10^{-3} $
$\Rightarrow \dfrac{n _2}{n _1}+1 =\dfrac{n _2+n _1}{n _1}=2.9 \times 10^{-3}+1 $
$\Rightarrow \dfrac{n _1}{n _1+n _2} =\chi _1=\dfrac{1}{1+2.9 \times 10^{-3}}=0.997 $
$\Rightarrow p =p _0 \chi _1=23.51 \times 0.997=23.44$ $ mm$
BETA
