Solve the following

Question:

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

Solution:

(0)

The balanced equation is

$\mathrm{SO}_{2}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{3}+\mathrm{H}_{2} \mathrm{O}$

moles of $\mathrm{NaOH}=$ molarity $\times$ volume (in litre)

$=0.1 \times 0.1$

$=0.01$ moles

Here $\mathrm{NaOH}$ is limiting Reagent 2 mole $\mathrm{NaOH} \longrightarrow 1$ mole $\mathrm{Na}_{2} \mathrm{SO}_{3}$

$0.01$ mole $\mathrm{NaOH} \longrightarrow \frac{1}{2} \times 0.01$ mole $\mathrm{Na}_{2} \mathrm{SO}_{3}$

Moles of $\mathrm{Na}_{2} \mathrm{SO}_{3} \longrightarrow 0.005$ mole

$\mathrm{Na}_{2} \mathrm{SO}_{3} \longrightarrow 2 \mathrm{Na}^{+}+\mathrm{SO}_{3}^{2-}$

$\mathrm{i}=3$

Moles of $\mathrm{H}_{2} \mathrm{O}=\frac{36}{18}=2$ moles

Accoding to RLVP -

$\frac{P_{A}^{0}-P_{A}}{P_{A}^{0}}=i X_{B}$

$\frac{P_{A}^{o}-P_{A}}{P_{A}^{0}}=\frac{i n_{A}}{i n_{A}+n_{B}}\left(i n_{A} \simeq 0\right)$

$\mathrm{n}_{\mathrm{B}}<<\mathrm{n}_{\mathrm{A}}$

$\left\{\boldsymbol{n}_{A}+\boldsymbol{n}_{B} \simeq \boldsymbol{n}_{A}\right\}$

$\frac{P_{A}^{o}-P_{A}}{P_{A}^{0}}=i \times \frac{n_{B}}{n_{A}}$

$\frac{2 \mathrm{H}-\mathrm{P}_{A}}{2 \mathrm{H}}=3 \times \frac{0.005}{2}$

$\Rightarrow 2 \mathrm{H}-\mathrm{P}_{\mathrm{A}}=0.18$

Lowering in pressure $=0.18 \mathrm{~mm}$ of $\mathrm{Hg}$ lowering in pressure $=18 \times 10^{-12} \mathrm{~mm}$ of $\mathrm{Hg}$

X =18

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