Question

$3.0$ molal $NaOH$ solution has a density of $1.110g/mL$ . The molarity of the solution is :

• A. $2.9732$
• B. $3.05$
• C. $3.64$
• D. $3.0504$

Answer 1

Answer: (A)

$NaOH$
$3$ molal
$d=1.11g/ml$
molality $=\frac{\text{ moles of solute }}{\text{ mass of solvent (kg) }}$
$3=\frac{n_{\text{solute }}}{1 kg}\begin{bmatrix} \because \text{ Let the mass of } \\ \text{ solvent }=1kg \end{bmatrix}$
$n_{\text{solute }}=3$
$\left(wt .\right)\_{solute}^{}=3\times 40$
$=120gm$
$M=\frac{\text{ moles }}{V \left(\right. L \left.\right)}=\frac{3}{1 . 009 \left(\right. L \left.\right)}\left[\begin{matrix} d=\frac{m}{v} \\ v=\frac{m \left(\right. solution \left.\right)}{d} \\ v=\frac{\left(\right. 1000 + 120 \left.\right)}{1 . 11} \\ v=1009mL \end{matrix}\right$
$\boxed{M = 2 . 9732}$