The correct relationship between molarity ( $M$ ) and molality $(m)$ is $(d =$ density of the solution, in ${KgL}^{- 1}, M_{2} =$ molar mass of the solute in $kg$ ${mol}^{- 1}$ )

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$M = \frac{md}{1 + {mM}_{2}}$

$M = \frac{m}{1 + {mM}_{2} d}$

$M = \frac{1 + {mM}_{2}}{md}$

$M = \frac{1 + md}{{mM}_{2}}$

answer is A.

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Detailed Solution

We know that

Molarity (M) $= \frac{N o . o f m o l e s o f s o l u t e}{V o l u m e o f s o l u t i o n {i n l i t r e s}}$

Molality (m) $= \frac{N o . o f m o l e s o f s o l u t e}{w e i g h t o f s o l v e n t}$

If a be density in $g / {cm}^{3}$ .

Then mass of the solution $= 1000 d gm$

mass of solute $= M \times M_{2}$

where $M_{2} =$ molar mass of solute

Mass of solvent $= 1000 d - {MXM}_{2}$

$∴ M o l a l i t y (m) = \frac{M \times 1000}{1000 d - M \times M_{2}}$

$∴ M = \frac{m d}{1 + m M_{2}}$

Hence, option (a) is the answer.

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