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

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Question

(C) Let the mass of solvent (water) be $1000 \ g$. The number of moles of water $(n_A)$ = $\frac{1000}{18} \ mol$. Let the number of moles of solute b

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the mass of solvent (water) be $1000 \ g$. The number of moles of water $(n_A)$ = $\frac{1000}{18} \ mol$. Let the number of moles of solute be $n_B = m$ (since molality is $m \ mol/kg$). The mole fraction of solute $(y)$ is given by $y = \frac{n_B}{n_A + n_B}$. Substituting the values: $y = \frac{m}{\frac{1000}{18} + m}$. Multiplying numerator and denominator by $18$: $y = \frac{18m}{1000 + 18m}$. Thus,the correct relation is $y = \frac{18m}{1000 + 18m}$.

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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