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(A) Given,mole fraction of water $X_{H_2O} = 0.85$. Therefore,mole fraction of $H_2SO_4$ is $X_{H_2SO_4} = 1 - 0.85 = 0.15$. Molality $(m)$ is defined

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$9.80$

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(A) Given,mole fraction of water $X_{H_2O} = 0.85$. Therefore,mole fraction of $H_2SO_4$ is $X_{H_2SO_4} = 1 - 0.85 = 0.15$. Molality $(m)$ is defined as the number of moles of solute per $1000 \ g$ of solvent. Moles of water $(n_{H_2O})$ in $1000 \ g$ of water = $\frac{1000 \ g}{18 \ g/mol} = 55.55 \ mol$. Using the relation $\frac{n_{H_2SO_4}}{n_{H_2O}} = \frac{X_{H_2SO_4}}{X_{H_2O}}$: $n_{H_2SO_4} = \frac{0.15}{0.85} 	imes 55.55 = 9.80 \ mol$. Since this amount of solute is present in $1000 \ g$ of solvent,the molality is $9.80 \ m$.

If the mole fraction of water in a solution of $H_2SO_4$ is $0.85$,then the molality of the solution is:

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