JEE Chemistry 2025 Question with Solution

Given: Nitric acid solution is $75\%$ by mass, density of solution is $1.25 \, \text{g/mL}$, and required mass of pure $\text{HNO}_3$ is $30 \, \text{g}$.

Find: The volume of solution containing $30 \, \text{g}$ of $\text{HNO}_3$.

Since the solution is $75\%$ by mass,

$$m_{\text{HNO}_3} = \frac{75}{100} \times m_{\text{total}}$$

Substituting $m_{\text{HNO}_3} = 30 \, \text{g}$,

$$30 = \frac{75}{100} \times m_{\text{total}}$$ $$m_{\text{total}} = \frac{30 \times 100}{75} = 40 \, \text{g}$$

Now use density:

$$\rho = \frac{m}{V}$$

So,

$$V = \frac{m}{\rho} = \frac{40}{1.25} = 32 \, \text{mL}$$

Therefore, the volume of the solution is $32 \, \text{mL}$.

The solution concludes $32 \, \text{mL}$, so the correct option is D. The answer key and listed "The Correct Option is C" disagree with the worked solution.

Given: Mass percentage of nitric acid = $75\%$ by mass, density of nitric acid solution = $1.25 \, \text{g/mL}$, mass of pure $\text{HNO}_3$ required = $30 \, \text{g}$.

Find: Volume of the solution.

A $75\%$ by mass solution means:

$$75 \, \text{g of HNO}_3 \text{ is present in } 100 \, \text{g of solution}$$

Therefore, for $30 \, \text{g}$ of $\text{HNO}_3$,

$$\text{Mass of solution} = \frac{30 \times 100}{75} = 40 \, \text{g}$$

Using the density relation,

$$\text{Density} = \frac{\text{Mass}}{\text{Volume}} \Rightarrow \text{Volume} = \frac{\text{Mass}}{\text{Density}}$$ $$\text{Volume} = \frac{40}{1.25} = 32 \, \text{mL}$$

Therefore, the required volume is $32 \, \text{mL}$, which corresponds to option D.