Question

$224mL$ of $\left(SO\right)_{2 \left(\right. g \left.\right)}$ 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) $\left( P _{\left( H _{2} O \right)}^{*}=24 mm\right.$ of $\left.Hg \right)$ is $x \times 10^{-2} mm$ of $Hg$, the value of $x$ is

Answer 1

The balanced equation is:
$SO_{2}+2NaOH \rightarrow Na_{2}SO_{3}+H_{2}O$
moles of $NaOH=$ molarity $\times $ volume (in litre)
$=0.1\times 0.1$
$=0.01$ moles
Here, $NaOH$ is limiting Reagent
$2$ mole $NaOH \rightarrow 1$ mole $Na_{2}SO_{3}$
$0.01$ mole $NaOH \rightarrow \frac{1}{2}\times 0.01moleNa_{2}SO_{3}$
Moles of $Na_{2}SO_{3} \rightarrow 0.005mole$
$Na_{2}SO_{3} \rightarrow 2Na^{+}+SO_{3}^{2 -}$
$i=3$
Moles of $H_{2}O=\frac{36}{18}=2$ moles
According to $RLVP$ -
$\frac{P_{A} ^\circ - P_{A}}{P_{A} ^\circ }=iX_{B}$
$\frac{P_{A}^{o} - P_{A}}{P_{A} ^\circ }=\frac{\left(in\right)_{A}}{\left(in\right)_{A} + n_{B}}\left(\left(in\right)_{A} \sim eq 0\right)$
$n_{B}>>n_{A}$
$\left\{n_{A} + n_{B} \sim eq n_{B}\right\}$
$\frac{P_{A}^{O} - P_{A}}{P_{A}^{0}}=i\times \frac{n_{B}}{n_{A}}$
$\frac{24 - P_{A}}{24}=\frac{3 \times 0 . 005}{2}$
$\Rightarrow 24-P_{A}=0.18$
Lowering in pressure $=0.18mm$ of $Hg$ lowering in pressure
$=18\times 10^{- 2}mm$ of $Hg$
$\boxed{x = 18}$