If the molality of an aqueous solution is m a | vedclass

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Question

(C) For an aqueous solution,the solvent is water $(M_1 = 18 \ g/mol)$.Molality $m$ is defined as $m = \frac{n_2 	imes 1000}{W_1}$,where $n_2$ is mole

Vedclass · Accepted Answer

$y = \frac{18m}{1000+18m}$

Vedclass · Answer

(C) For an aqueous solution,the solvent is water $(M_1 = 18 \ g/mol)$.Molality $m$ is defined as $m = \frac{n_2 	imes 1000}{W_1}$,where $n_2$ is moles of solute and $W_1$ is mass of solvent in grams.Mole fraction of solute $y = \frac{n_2}{n_1 + n_2}$.Since $n_1 = \frac{W_1}{18}$,we have $W_1 = 18n_1$.Substituting $W_1$ in the molality formula: $m = \frac{n_2 	imes 1000}{18n_1} = \frac{1000}{18} 	imes \frac{n_2}{n_1}$.From the mole fraction definition,$\frac{n_2}{n_1} = \frac{y}{1-y}$.Therefore,$m = \frac{1000}{18} 	imes \frac{y}{1-y}$.Rearranging for $y$: $18m(1-y) = 1000y \implies 18m - 18my = 1000y \implies 18m = y(1000 + 18m)$.Thus,$y = \frac{18m}{1000 + 18m}$.

If the molality of an aqueous solution is $m$ and the mole fraction of the solute is $y$,which of the following relations is correct?

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