The vapour pressure of an aqueous solution is found to be $750 torr$ at certain temperature $^{'} T^{'}$ . If $^{'} T^{'}$ is the temperature at which pure water boils under atmospheric pressure and same solution show elevation in boiling point $Δ T_{b} = 1.04 K$ , find the atmospheric pressure $(K_{b} = 0.52 {k kg mol}^{- 1})$

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Solution

$Elevation in boiling point Δ T_{b} = k_{b} \times m$
$W h e r e, m = molality$
$On substituting given values,$
we get $m = \frac{Δ T_{b}}{k_{b}} = \frac{1.04}{0.52}$
$Molality (m) = 2 mol/kg$
So, $Moles of solute (n_{solute}) = 2 mol$
$Moles of solvent (n_{H_{2} O}) = \frac{1000}{18} = 55.55 mol$
The vapour pressure of the aqueous solution is given as $750 torr$
$According to Raoult'slaw:$
$\frac{P^{\circ} - P_{solution}}{P_{solution}} = \frac{n_{solute}}{n_{solvent}}$
$\Rightarrow \frac{P^{\circ} - 750}{750} = \frac{2}{\frac{1000}{18}}$
$\Rightarrow P^{\circ} = \frac{750 \times 2}{\frac{1000}{18}} + 750$
$\Rightarrow P^{\circ} = 27 + 750$
$\Rightarrow P^{\circ} = 777 torr$

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The vapour pressure of an aqueous solution is found to be 750 torr at certain temperature ′T′. If ′T′ is the temperature at which pure water boils under atmospheric pressure and same solution show elevation in boiling point ΔTb=1.04 K, find the atmospheric pressure (Kb=0.52 k kg mol−1)