224 mL of SO_2(g) at 298 K and 1 atm is passed through 100 mL of 0.1 M NaOH solution. The non-volatile solute produced is dissolved in 36g of water. The lowering of vapour pressure of solution (assuming the solution in dilute) (P$_{({H_2}O)}^o$ $-$ 24 mm of Hg) is x $\times$ 10^$-$2 mm of Hg, the value of x is ___________. (Integer answer)

moles of SO₂ = ${{224} \over {22400}}$ = 0.01 <br><br>moles of NaOH = molarity × volume (in litre) <br>= 0.1 × 0.1 <br>= 0.01 moles <br><br>The balanced equation is <br><br>SO₂ + 2NaOH $\to$ Na₂SO₃ + H₂O <br><br>$\therefore$ Here NaOH is limiting Reagent. <br><br>2 mole NaOH produces 1 mole Na₂SO₃ <br><br>0.01 mole NaOH produces ${1 \over 2}$ $\times$ 0.01 mole Na₂SO₃ <br><br>$\therefore$ Moles of Na₂SO₃ = 0.005 mole <br><br>Na₂SO₃ $\to$ 2Na⁺ + SO₃^2- <br><br>van’t Hoff factor (i) = 3 <br><br>Moles of H₂O = ${{36} \over {18}}$ = 2 moles <br><br>Accoding to Relative Lowering of Vapour : <br><br>$${{P_{{H_2}O}^o - {P_S}} \over {P_{{H_2}O}^o}} = {{i{n_{N{a_2}C{O_3}}}} \over {{n_{{H_2}O}} + i{n_{N{a_2}C{O_3}}}}}$$ <br><br>[ ${{n_{N{a_2}C{O_3}}}}$ &lt;&lt; ${{n_{{H_2}O}}}$ ] <br><br>$${{P_{{H_2}O}^o - {P_S}} \over {P_{{H_2}O}^o}} = {{i{n_{N{a_2}C{O_3}}}} \over {{n_{{H_2}O}}}}$$ <br><br>$\Rightarrow$ ${{24 - {P_S}} \over {24}} = {{3 \times 0.005} \over 2}$ <br><br>$\Rightarrow$ 24 – P_S = 0.18 <br><br>$\Rightarrow$ P_S = 23.82 <br><br>Lowering in pressure ($\Delta$P) = 0.18 mm of Hg <br><br>= 18 × 10^–2 mm of Hg