Question

Density of $3M$ solution $NaCl$ is $1.25g/cc$ . The molality of solution is

• A. $2.79$ molal
• B. $0.279$ molal
• C. $1.279$ molal
• D. $3.85$ molal

Answer 1

Answer: (A)

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
$n_{NaCl}=3$ mol
Mass of $1L$ solution of $NaCl=($ volume $)($ density $)$
$=(1000mL)\left(1.25gmo{l}^{-1}\right)$
$=1250g$
Mass of $NaCl$ in $1L$ of solution $=\left(n_{NaCl}\right)\left(m{m}_{NaCl}\right)$
$=(3mol)\left(58.5gmo{l}^{-1}\right)$
$=175.5g$
Mass of solvent $\left(H_{2}O\right)=$ Mass of solution - Mass of solute $=(1250g)-(175.5g)$
$=1074.5g$
Thus
$m=\frac{n_{\text{solute }}}{g_{\text{solv }}\times }\frac{1000g}{kg}$
$=\frac{3\text{ mol }}{1074.5g}\times \frac{1000g}{kg}$
$2.79$ molkg ${}^{-1}$
Alternatively $m=\frac{M_{\text{soln }}}{1000d-M_{\text{soln }m{m}_{\text{solute }}}}\times 1000$
$=\frac{3}{(1000)(1.25)-(3)(58.5)}\times 1000$
$=2.79molk{g}^{-1}$