Solutions and Colligative Properties - Result Question 16-1
16. For a dilute solution containing $2.5$ g of a non-volatile non-electrolyte solute in $100$ g of water, the elevation in boiling point at $1$ atm pressure is $2^{\circ} \mathrm{C}$. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure ( mm of Hg ) of the solution is (take $K_b=0.76$ $\mathrm{K} $ $\mathrm{kg}$ $ \mathrm{mol}^{-1}$ ).
(2012)
(a) $724$
(b) $740$
(c) $736$
(d) $ 718$
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Answer:
Correct Answer: 16. (a)
Solution:
The elevation in boiling point is
$ \Delta T_b=K_b \cdot m: m=\text { molality }=\frac{n_2}{w_1} \times 1000 $
$\left[n_2=\right.$ Number of moles of solute, $w_1=$ Weight of solvent in gram]
$ \begin{array}{ll} \Rightarrow & 2=0.76 \times \frac{n_2}{100} \times 1000 \\ \Rightarrow & n_2=\frac{5}{19} \end{array} $
Also, from Raoult’s law of lowering of vapour pressure :
$\frac{-\Delta p}{p^{\circ}} =x_2=\frac{n_2}{n_1+n_2} \approx \frac{n_2}{n_1} \quad\left[\because n_1 \gg n_2\right]$
$\Rightarrow \quad-\Delta p =760 \times \frac{5}{19} \times \frac{18}{100}=36 $ $\mathrm{mm} \text { of } \mathrm{Hg}$
$\Rightarrow \quad p =760-36=724 $ $\mathrm{mm} \text { of } \mathrm{Hg} $
BETA
