Solutions and Colligative Properties - Result Question 24
23. The vapour pressure of a dilute aqueous solution of glucose $\left(C _6 H _{12} O _6\right)$ is $750$ $ mm$ of mercury at $373 K$. Calculate (i) molality and (ii) mole fraction of the solute.
$(1989,3 M)$
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Answer:
Correct Answer: 23. $(0.75)$
Solution:
At $373$ $ K$ (bp) of $H _2 O$, Vapour pressure $=760$ $ mm$
VP of solution at $373 K=750$ $ mm$
$ \Rightarrow \quad p=p _0 \chi _1 $
or $\quad 750=760 \chi _1$
$\Rightarrow \quad \chi _1=\dfrac{75}{76}=$ mole fraction of $H _2 O$
$\Rightarrow \quad \chi _2=1-\dfrac{75}{76}=\dfrac{1}{76}=$ mole fraction of solute
Now $\quad \dfrac{n _2}{n _1+n _2}=\dfrac{1}{76}$
$\Rightarrow \quad \dfrac{n _1}{n _2}=75$
$\Rightarrow \quad$ Molality $=\dfrac{n _2}{n _1 M _1} \times 1000=\dfrac{1000}{75 \times 18}=0.74$ molal
BETA
