224 , mL of SO_{2(g)} at 273 , K and 1 , atm | vedclass

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(B) Step $1$: Calculate moles of reactants.$n_{SO_2} = \frac{224 \, mL}{22400 \, mL/mol} = 0.01 \, mol$.$n_{NaOH} = 0.1 \, M 	imes 0.1 \, L = 0.01 \,

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) Step $1$: Calculate moles of reactants.$n_{SO_2} = \frac{224 \, mL}{22400 \, mL/mol} = 0.01 \, mol$.$n_{NaOH} = 0.1 \, M 	imes 0.1 \, L = 0.01 \, mol$.Step $2$: Reaction stoichiometry.$SO_2 + NaOH ightarrow NaHSO_3$.Since $n_{SO_2} = n_{NaOH} = 0.01 \, mol$, both are consumed to produce $0.01 \, mol$ of $NaHSO_3$.Step $3$: Calculate lowering of vapour pressure.$\Delta P = P^0_{H_2O} 	imes \chi_{solute} = P^0_{H_2O} 	imes \frac{n_{solute}}{n_{solute} + n_{solvent}}$.$n_{solvent} = \frac{36 \, g}{18 \, g/mol} = 2 \, mol$.$\Delta P = 24 	imes \frac{0.01}{0.01 + 2} = 24 	imes \frac{0.01}{2.01} \approx 24 	imes 0.004975 = 0.1194 \, mm \, Hg$.$\Delta P = 11.94 	imes 10^{-2} \, mm \, Hg$.Rounding to the nearest integer, $x = 12$.

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