Question

The vapour pressure of an aqueous solution is found to be $750\text{torr}$ at certain temperature ${}^{\prime }{T}^{\prime }$. If ${}^{\prime }{T}^{\prime }$ is the temperature at which pure water boils under atmospheric pressure and same solution show elevation in boiling point $\Delta T_b=1.04\text{K}$, then let the atmospheric pressure be $x$. Find $\frac{x}{111}$ $(K_b=0.52{\text{k kg mol}}^{-1})$

• A. 7.00
• B. 7
• C. 7.0

Answer 1

Answer: (A), (B), (C)

$\text{Elevation in boiling point}\Delta T_b=k_b\times m$
$Where,m=\text{molality}$
$\text{On substituting given values,}$
we get $m=\frac{\Delta T_b}{k_b}=\frac{1.04}{0.52}$
$\text{Molality}\left(m\right)=2\text{mol/kg}$
So, $\text{Moles of solute}\left(n_{\text{solute}}\right)=2\text{mol}$
$\text{Moles of solvent}\left(n_{H_{2}O}\right)=\frac{1000}{18}=55.55\text{mol}$
The vapour pressure of the aqueous solution is given as $750\text{torr}$
$\text{According to Raoult'slaw:}$
$\frac{P^{\circ }-P_{\text{solution}}}{P_{\text{solution}}}=\frac{n_{\text{solute}}}{n_{\text{solvent}}}$
$\Rightarrow \frac{P^{\circ }-750}{750}=\frac{2}{\frac{1000}{18}}$
$\Rightarrow P^{\circ }=\frac{750\times 2}{\frac{1000}{18}}+750$
$\Rightarrow P^{\circ }=27+750$
$\Rightarrow P^{\circ }=x=777\text{torr}$
$\therefore \frac{x}{111}=\frac{777}{111}=7$