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 ? $$($$ $$dH2O$$ $$=$$ $$1$$ g cm–$$3)$$

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An amount of $$20g$$ . . . . . .Before electrolysis mass (H2O) $$=$$ $$150$$ gAfter $$electrolysismH2O=150$$−$$113.3=36.7gnH2O=2.04molSince$$,$$2H2O$$→$$2H2+O2$$    $$nH2=2.04mol$$                         $$V=45.6LnO2=1.02mol$$.                     $$V=22.8L$$

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)

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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 ? ( dH2O = 1 g cm–3)

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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)