If the mole fraction of a solute in liquid na | vedclass

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(C) The molar mass of naphthalene $(C_{10}H_8)$ is $M_{solvent} = (10 	imes 12) + (8 	imes 1) = 128 \ g/mol$. Given: Mole fraction of solute $(x_2)

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$1.95$

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(C) The molar mass of naphthalene $(C_{10}H_8)$ is $M_{solvent} = (10 	imes 12) + (8 	imes 1) = 128 \ g/mol$. Given: Mole fraction of solute $(x_2) = 0.2$. Therefore,mole fraction of solvent $(x_1) = 1 - 0.2 = 0.8$. Molality $(m)$ is given by the formula: $m = \frac{x_2 	imes 1000}{x_1 	imes M_{solvent}}$. Substituting the values: $m = \frac{0.2 	imes 1000}{0.8 	imes 128}$. $m = \frac{200}{102.4} \approx 1.95 \ m$.

If the mole fraction of a solute in liquid naphthalene $(C_{10}H_8)$ is $0.2$,then the molality of the solution will be .............. $m$.

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