The expression relating mole fraction of solute $(χ_{2})$ and molarity (M) of the solution is (where d is the density of the solution in $g L^{- 1}$ and $M w_{1}$ and $M w_{2}$ are the molar masses of solvent and solute, respectively) :

x_{2} = \frac{M \times M w_{1}}{M (M w_{1} - M w_{2}) + 1000 d}

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x_{2} = \frac{M \times M w_{1}}{M (M w_{1} - M w_{2}) + d}

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x_{2} = \frac{M \times M w_{1}}{M (M w_{1} - M w_{2}) - 1000 d}

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x_{2} = \frac{M \times M w_{1}}{M (M w_{1} - M w_{2}) - d}

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Solution

The correct option is A $x_{2} = \frac{M \times M w_{1}}{M (M w_{1} - M w_{2}) + d}$
Let, the volume of solution $= 1 L$
Weight of solution $= 1 \times d = d g$
Number of moles of solute $(n_{2})$ in 1 L solution $= M$
$∴ \frac{W_{2}}{M w_{2}} = M$
$W_{2}$ (weight of solute) $= M \times M w_{2}$
$W_{1}$ (weight of solvent) $=$ Weight of solution - Weight of solute
$= (d - M \times M w_{2})$
Number of moles of solvent $(n_{1})$
$= \frac{W_{1}}{M w_{1}} = (\frac{d - M \times M w_{2}}{M w_{1}})$
$∴ χ_{2} = \frac{n_{2}}{n_{1} + n_{2}} = \frac{M}{⎛ ⎝ \frac{d - M \times M w_{2}}{M w_{1}} ⎞ ⎠ + m}$
$= \frac{M \times M w}{M) M w_{1} - M w_{2}) + d}$

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