The mole fraction of water in a 20% aqueous s | vedclass

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(B) $20\%$ aqueous solution of $H_2O_2$ by mass means $20 \ g$ of $H_2O_2$ is present in $100 \ g$ of the solution.Therefore,the mass of water = $100

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$\frac{68}{77}$

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(B) $20\%$ aqueous solution of $H_2O_2$ by mass means $20 \ g$ of $H_2O_2$ is present in $100 \ g$ of the solution.Therefore,the mass of water = $100 \ g - 20 \ g = 80 \ g$.Moles of $H_2O_2$ $(n_{H_2O_2})$ = $\frac{20 \ g}{34 \ g/mol} = \frac{20}{34} \ mol$.Moles of $H_2O$ $(n_{H_2O})$ = $\frac{80 \ g}{18 \ g/mol} = \frac{80}{18} \ mol$.Mole fraction of water $(x_{H_2O})$ = $\frac{n_{H_2O}}{n_{H_2O} + n_{H_2O_2}} = \frac{\frac{80}{18}}{\frac{80}{18} + \frac{20}{34}}$.$x_{H_2O} = \frac{4.44}{4.44 + 0.588} = \frac{4.44}{5.028} \approx 0.883$.Calculating the fraction: $\frac{80/18}{(80 	imes 34 + 20 	imes 18) / (18 	imes 34)} = \frac{80 	imes 34}{2720 + 360} = \frac{2720}{3080} = \frac{272}{308} = \frac{68}{77}$.

The mole fraction of water in a $20\%$ aqueous solution of $H_2O_2$ is:

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