Question

$224mL$ of ${\left(SO\right)}_{2(g)}$ at $298K$ and $1atm$ is passed through $100mL$ of $0.1MNaOH$ solution. The non-volatile solute produced is dissolved in $36g$ of water. The lowering of vapour pressure of solution (assuming the solution is dilute) $(P_{\left(H_{2}O\right)}^{\ast }=24mm$ of $Hg)$ is $x\times 10^{-2}mm$ of $Hg$, the value of $x$ is

Answer 1

Answer: 18

The balanced equation is:
$S{O}_2+2NaOH\to N{a}_{2}S{O}_3+H_{2}O$
moles of $NaOH=$ molarity $\times$ volume (in litre)
$=0.1\times 0.1$
$=0.01$ moles
Here, $NaOH$ is limiting Reagent
$2$ mole $NaOH\to 1$ mole $N{a}_{2}S{O}_3$
$0.01$ mole $NaOH\to \frac{1}{2}\times 0.01moleN{a}_{2}S{O}_3$
Moles of $N{a}_{2}S{O}_3\to 0.005mole$
$N{a}_{2}S{O}_3\to 2N{a}^{+}+S{O}_3^{2-}$
$i=3$
Moles of $H_{2}O=\frac{36}{18}=2$ moles
According to $RLVP$ -
$\frac{P_A^{\circ }-P_A}{P_A^{\circ }}=i{X}_B$
$\frac{P_A^o-P_A}{P_A^{\circ }}=\frac{{\left(in\right)}_A}{{\left(in\right)}_A+n_B}\left({\left(in\right)}_A\sim eq0\right)$
$n_B>>n_A$
$\left\{n_A+n_B\sim eq{n}_B\right\}$
$\frac{P_A^O-P_A}{P_A^0}=i\times \frac{n_B}{n_A}$
$\frac{24-P_A}{24}=\frac{3\times 0.005}{2}$
$\Rightarrow 24-P_A=0.18$
Lowering in pressure $=0.18mm$ of $Hg$ lowering in pressure
$=18\times 10^{-2}mm$ of $Hg$
$\boxed{x=18}$