Solutions and Colligative Properties - Result Question 52
####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.
$(1989,3 M)$
Show Answer
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 $\quad 750=760 \chi _1$
$\Rightarrow \quad \chi _1=\frac{75}{76}=$ mole fraction of $H _2 O$
$\Rightarrow \quad \chi _2=1-\frac{75}{76}=\frac{1}{76}=$ mole fraction of solute
Now $\quad \frac{n _2}{n _1+n _2}=\frac{1}{76}$
$\Rightarrow \quad \frac{n _1}{n _2}=75$
$\Rightarrow \quad$ Molality $=\frac{n _2}{n _1 M _1} \times 1000=\frac{1000}{75 \times 18}=0.74$ molal
