$50 \mathrm{~g}$ of a substance is dissolved in $1 \mathrm{~kg}$ of water at $+90^{\circ} \mathrm{C}$. The…
$50 \mathrm{~g}$ of a substance is dissolved in $1 \mathrm{~kg}$ of water at $+90^{\circ} \mathrm{C}$. The temperature is reduced to $+10^{\circ} \mathrm{C}$. The density is increased from 1.1 to $1.15 \mathrm{~g} \mathrm{cc}^{-1}$. What is the $\%$ change of molarity of the solution?
$10$
$4.5$
$5$
$7.3$
Solution
Molarity $=\frac{\text { Weight dissolved }}{\text { Molar mass }} \times \frac{1000}{V(\mathrm{~mL})}$ and
$V=\frac{\text { Mass of solution }}{\text { Density }}$
Molarity (1) at $90^{\circ} \mathrm{C}=\frac{50}{x} \times \frac{1000 \times 1.1}{1050}=\frac{52.38}{x}$
Molarity (2) at $10^{\circ} \mathrm{C}=\frac{50}{x} \times \frac{1000 \times 1.15}{1050}=\frac{54.76}{x}$
Change in molarity $=$ Molarity (2) - Molarity (1) $\%$ change in molarity will be
$=\frac{\text { Molarity }(2)-\text { Molarity }(1)}{\text { Molarity (1) }} \times 100$
$=\frac{\frac{54.76}{x}-\frac{52.38}{x}}{\frac{52.38}{x}} \times 100$
$=\frac{2.38}{x} \times \frac{x}{52.38} \times 100=4.5$