Question

224 mL of \(\mathrm{SO_{2(g)}}\) 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) (\(\mathrm{P^o_{(H_2O)}}\) - 24 mm of Hg) is x \(\times\) \(10^{-2}\) mm of Hg, the value of x is ___________. (Integer answer)

Hint:

Find the moles of \(\mathrm{SO_2}\) and NaOH, work out the salts formed, count dissolved particles (van't Hoff factor), then use \(\Delta P = P^o\,\dfrac{n_{solute}}{n_{water}}\).

Answer 1

Correct Answer: 23

\(n_{SO_2} = \dfrac{PV}{RT} = \dfrac{1 \times 0.224}{0.0821 \times 298} = 0.00916\) mol; \(n_{NaOH} = 0.1 \times 0.1 = 0.01\) mol.
First \(\mathrm{SO_2 + 2NaOH \to Na_2SO_3 + H_2O}\) uses 0.005 mol \(\mathrm{SO_2}\) and all the NaOH, giving 0.005 mol \(\mathrm{Na_2SO_3}\). The remaining 0.00416 mol \(\mathrm{SO_2}\) reacts: \(\mathrm{Na_2SO_3 + SO_2 + H_2O \to 2NaHSO_3}\).
Final: \(\mathrm{Na_2SO_3}\) = 0.00084 mol, \(\mathrm{NaHSO_3}\) = 0.00832 mol. Dissolved particles: \(\mathrm{Na^+}\) = 0.01, \(\mathrm{SO_3^{2-}}\) = 0.00084, \(\mathrm{HSO_3^-}\) = 0.00832, total = 0.0192 mol.
\(n_{H_2O} = 36/18 = 2\) mol. For the dilute solution, \(\Delta P = P^o \times \dfrac{0.0192}{2} = 24 \times 0.0096 = 0.23\) mm Hg \(= 23 \times 10^{-2}\) mm Hg.
So x = 23 (the NTA key accepts a range 18 to 24: the intended 0.01 mol \(\mathrm{SO_2}\) gives 24 with \(\mathrm{NaHSO_3}\), or 18 with \(\mathrm{Na_2SO_3}\)).
Answer: 23