The density of 3 text{ M} solution of sodium | vedclass

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(C) The relation between molarity $(M)$ and molality $(m)$ is given by the formula: $d = M \left( \frac{1}{m} + \frac{M_2}{1000} \right)$,where $d$ is

Vedclass · Accepted Answer

$2.79$

Vedclass · Answer

(C) The relation between molarity $(M)$ and molality $(m)$ is given by the formula: $d = M \left( \frac{1}{m} + \frac{M_2}{1000} ight)$,where $d$ is density in $	ext{g mL}^{-1}$,$M$ is molarity,$m$ is molality,and $M_2$ is the molar mass of the solute.Given: $M = 3 	ext{ M}$,$d = 1.252 	ext{ g mL}^{-1}$,$M_2 = 58.5 	ext{ g mol}^{-1}$.Substituting the values: $1.252 = 3 \left( \frac{1}{m} + \frac{58.5}{1000} ight)$.$1.252 / 3 = \frac{1}{m} + 0.0585$.$0.41733 = \frac{1}{m} + 0.0585$.$\frac{1}{m} = 0.41733 - 0.0585 = 0.35883$.$m = 1 / 0.35883 \approx 2.79 	ext{ m}$.

The density of $3 \text{ M}$ solution of sodium chloride is $1.252 \text{ g mL}^{-1}$. The molality of the solution will be ............. $\text{m}$. (molar mass,$NaCl = 58.5 \text{ g mol}^{-1}$)

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