Question

50 g of a substance is dissolved in 1 kg of water at $ 90^\circ C $. When the temperature is lowered to $ 10^\circ C $, the density increases from $ 1.1\, \text{g/cc} $ to $ 1.15\, \text{g/cc} $. What is the percentage change in molarity of the solution?

• A. 10
• B. 4.5
• C. 5
• D. 7.3

Hint:

Molarity changes with volume. Use density to find volume and compute percentage change: \( \Delta M\% = \frac{\Delta V}{V} \times 100 \)

Answer 1

The Correct Option is B

Let’s assume: - Mass of solution = \( 50 + 1000 = 1050\, \text{g} \) At \( 90^\circ C \), density \( \rho_1 = 1.1\, \text{g/cc} \Rightarrow V_1 = \frac{1050}{1.1} = 954.5\, \text{cc} \) At \( 10^\circ C \), density \( \rho_2 = 1.15\, \text{g/cc} \Rightarrow V_2 = \frac{1050}{1.15} = 913\, \text{cc} \) Change in molarity is inversely proportional to volume: \[ \text{% change} = \left( \frac{V_1 - V_2}{V_1} \right) \times 100 = \left( \frac{954.5 - 913}{954.5} \right) \times 100 \approx 4.36% \approx 4.5% \]