An amount of 20g potassium sulphate was dissolved in 150 cm³ of water. The solution was then electrolysed. After electrolysis the content of potassium sulphate in the solution was 15% by mass. Determine volume of the gases obtained at NTP ? ( d_H2O = 1 g cm^–3)

On electrolysis, only water is decomposed and the total amount of potassium sulphate in the electrolyte solution is constant. The mass of water in the solution: 

Before electrolysis mass (H₂O) = 150 g

After electrolysis:

$$\begin{gathered} \mathrm{m}\left({\mathrm{H}}_2\mathrm{O}\right)=\mathrm{m}\left(\mathrm{solution}\right)-\mathrm{m}\left({\mathrm{K}}_2{\mathrm{SO}}_4\right) \\ =20\times \frac{15}{100}-20=113.3 \mathrm{g} \end{gathered}$$

The mass of water decomposed on electrolysis:

$$\begin{gathered} \mathrm{m}\left({\mathrm{H}}_2\mathrm{O}\right)=150 \mathrm{g}-113.3 \mathrm{g}=36.7 \mathrm{g} \\ \mathrm{moles}=\frac{\mathrm{mass}}{\mathrm{molar} \mathrm{mass}} \\ \mathrm{moles}=\frac{36.7 \mathrm{g}}{18 \mathrm{g}/\mathrm{mol}}=2.04 \mathrm{mol} \end{gathered}$$

${\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}=2.04\mathrm{mol}$

Since,

$\begin{array}{l}2{\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{H}}_2+{\mathrm{O}}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\\ {\mathrm{n}}_{{\mathrm{H}}_2}=2.04 \mathrm{mol}\hspace{0.25em}\left(2\times 1.02\right)\\ {\mathrm{n}}_{{\mathrm{O}}_2}=1.02 \mathrm{mol}\hspace{0.25em} \end{array}$

$$\begin{gathered} \mathrm{PV}=\mathrm{nRT} \\ {\mathrm{V}}_{{\mathrm{H}}_2}=\frac{\mathrm{nRT}}{\mathrm{P}} \\ {\mathrm{V}}_{{\mathrm{H}}_2}=\frac{2.04 \mathrm{mol}\times 0.08205 \mathrm{L}.\mathrm{atm}.{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}\times 293.15 \mathrm{K}}{1 \mathrm{atm}}=49 \mathrm{L} \end{gathered}$$

$$\begin{gathered} \mathrm{PV}=\mathrm{nRT} \\ {\mathrm{V}}_{{\mathrm{O}}_2}=\frac{\mathrm{nRT}}{\mathrm{P}} \\ {\mathrm{V}}_{{\mathrm{O}}_2}=\frac{1.02 \mathrm{mol}\times 0.08205 \mathrm{L}.\mathrm{atm}.{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}\times 293.15 \mathrm{K}}{1 \mathrm{atm}}=24.53 \mathrm{L} \end{gathered}$$