# Question #cd313

specific gravity of the solution at a particular temperature $\left({40}^{\circ} C\right)$
$s = \text{mass of the solution up to etched mark"/"mass of equal volume water}$
$= \frac{\left(20.3300 - 20.23102\right) g}{\left(20.2376 - 20.23102\right) g} = \frac{0.09898}{0.00658} \approx 15$
So density of the solution $d = 15 g \text{/} m L$ at ${40}^{\circ} C$