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: Recall the definition of molarity.
Molarity is given by \[ M=\frac{\text{number of moles of solute}}{\text{volume of solution in litres}} \] Since the amount of solute remains constant, molarity changes only due to change in volume.

Step 2: Relate volume with density.
Density is given by \[ \rho=\frac{m}{V} \] Therefore, \[ V=\frac{m}{\rho} \] Since total mass of the solution remains constant, \[ M\propto \rho \] Thus, \[ \frac{M_2}{M_1}=\frac{\rho_2}{\rho_1} \]

Step 3: Substitute the densities.
Given, \[ \rho_1=1.1\,\text{g cc}^{-1} \] and \[ \rho_2=1.15\,\text{g cc}^{-1} \] Therefore, \[ \frac{M_2}{M_1}=\frac{1.15}{1.1} \] \[ \frac{M_2}{M_1}=1.04545 \]

Step 4: Calculate percentage change in molarity.
\[ \%\text{ change} = \left( \frac{M_2-M_1}{M_1} \right)\times100 \] \[ =(1.04545-1)\times100 \] \[ =0.04545\times100 \] \[ \approx4.5\% \]

Step 5: Final conclusion.
Hence, the percentage change in molarity is \[ \boxed{4.5\%} \]