In a 20% aqueous solution of H_2O_2,the mole | vedclass

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(B) $20\%$ $H_2O_2$ by mass means $20 \ g$ of $H_2O_2$ in $100 \ g$ of solution. Mass of $H_2O = 100 \ g - 20 \ g = 80 \ g$. Moles of $H_2O_2 (n_1) =

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$\frac{68}{77}$

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(B) $20\%$ $H_2O_2$ by mass means $20 \ g$ of $H_2O_2$ in $100 \ g$ of solution. Mass of $H_2O = 100 \ g - 20 \ g = 80 \ g$. Moles of $H_2O_2 (n_1) = \frac{20}{34} \approx 0.588 \ mol$. Moles of $H_2O (n_2) = \frac{80}{18} \approx 4.444 \ mol$. Total moles $(n_{total}) = n_1 + n_2 = \frac{20}{34} + \frac{80}{18} = \frac{10}{17} + \frac{40}{9} = \frac{90 + 680}{153} = \frac{770}{153}$. Mole fraction of water $(X_{H_2O}) = \frac{n_2}{n_{total}} = \frac{80/18}{770/153} = \frac{80}{18} 	imes \frac{153}{770} = \frac{40}{9} 	imes \frac{153}{770} = \frac{40 	imes 17}{770} = \frac{680}{770} = \frac{68}{77}$.

In a $20\%$ aqueous solution of $H_2O_2$,the mole fraction of water is $.......$.

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