If the mole fraction of H_2SO_4 in an aqueous | vedclass

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(A) Given: Mole fraction of $H_2SO_4$ $(x_2)$ = $0.1$. Since it is an aqueous solution,the mole fraction of water $(x_1)$ = $1 - 0.1 = 0.9$. Molality

Vedclass · Accepted Answer

$6.17$

Vedclass · Answer

(A) Given: Mole fraction of $H_2SO_4$ $(x_2)$ = $0.1$. Since it is an aqueous solution,the mole fraction of water $(x_1)$ = $1 - 0.1 = 0.9$. Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent. $m = \frac{n_2}{W_1 (	ext{in kg})} = \frac{n_2 	imes 1000}{n_1 	imes M_1}$,where $n_2$ is moles of solute,$n_1$ is moles of solvent,and $M_1$ is the molar mass of solvent (water = $18 \ g/mol$). $m = \frac{x_2 	imes 1000}{x_1 	imes M_1} = \frac{0.1 	imes 1000}{0.9 	imes 18} = \frac{100}{16.2} \approx 6.17 \ m$.

If the mole fraction of $H_2SO_4$ in an aqueous solution of $H_2SO_4$ is $0.1$,then the molality of the solution is ........... $m$.

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