Question

\(50\,\text{g}\) of a substance is dissolved in \(1\,\text{kg}\) of water at \(+90^\circ\text{C}\). The temperature is reduced to \(+10^\circ\text{C}\). The density is increased from \(1.1\) to \(1.15\,\text{g cc}^{-1}\). What is the % change of molarity of the solution?

• A. \(10\)
• B. \(4.5\)
• C. \(5\)
• D. \(7.3\)

Hint:

For a solution of constant mass, \[ M\propto \rho \] because molarity depends inversely on volume and \[ V=\frac{m}{\rho} \]

Answer 1

The Correct Option is B

Step 1: Understand the relationship between molarity and density.
Molarity $M = n/V$. Since moles of solute $n$ are unchanged when temperature changes, molarity changes only due to change in solution volume.
Step 2: Relate volume to density at constant total mass.
Total mass = $50 + 1000 = 1050$ g (constant). Since $V = m/\rho$: \[ M \propto \frac{1}{V} \propto \rho \quad \Rightarrow \quad \frac{M_2}{M_1} = \frac{\rho_2}{\rho_1} \]
Step 3: Identify densities at each temperature.
At $90^\circ C$: $\rho_1 = 1.1$ g/cc. At $10^\circ C$: $\rho_2 = 1.15$ g/cc.
Step 4: Calculate percentage change in molarity.
\[ \% \text{ change} = \left(\frac{\rho_2}{\rho_1} - 1\right) \times 100 = \left(\frac{1.15}{1.1} - 1\right) \times 100 = \frac{0.05}{1.1} \times 100 \approx 4.5\% \]
Step 5: Interpret the result physically.
As temperature decreases, density increases, volume decreases, molarity increases by ~4.5%. This is physically reasonable for a 0.05 g/cc density rise.
Step 6: State the final answer.
\[ \boxed{4.5\%} \]