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Q.
Density of $ 3\,M\, $ solution $NaCl $ is $ 1.25 \,g /cc $ . The molality of solution is
AMUAMU 2015Solutions
Solution:
Molality $(m)$ is given as
$m=\frac{n_{\text {solute }}}{k g_{\text {solvent }}}$
To calculate no. of moles of solute we take the help of molarity and to determine mass of solvent we take the help of density of solution.
$3M$ solution of $NaCl$ implies that there are $3$ moles of $NaCl$ in every 1 litre of solution. Considering $1L$ of solution, we have
$n_{N a C l}=3$ mol
Mass of $1 L$ solution of $NaCl =($ volume $)($ density $)$
$=(1000 m L)\left(1.25 g m o l^{-1}\right)$
$=1250 g$
Mass of $NaCl$ in $1 L$ of solution $=\left(n_{N a C l}\right)\left(m m_{N a C l}\right)$
$=(3 m o l)\left(58.5 g m o l^{-1}\right)$
$=175.5 g$
Mass of solvent $\left(H_{2} O\right)=$ Mass of solution - Mass of solute $=(1250 g)-(175.5 g)$
$=1074.5 g$
Thus
$m=\frac{n_{\text {solute }}}{g_{\text {solv }} \times} \frac{1000 g}{k g}$
$=\frac{3 \quad \text { mol }}{1074.5 g} \times \frac{1000 g}{k g}$
$2.79$ molkg $^{-1}$
Alternatively $m=\frac{M_{\text {soln }}}{1000 d-M_{\text {soln } m m_{\text {solute }}}} \times 1000$
$=\frac{3}{(1000)(1.25)-(3)(58.5)} \times 1000$
$=2.79 molkg ^{-1}$