JEE Chemistry 2023 Question with Solution

Given:

• Monobasic strong acid with molar mass $24.2 \, \text{g mol}^{-1}$
• Density of acid solution = $1.21 \, \text{kg L}^{-1}$
• NaOH volume = $25 \, \text{mL}$
• NaOH concentration = $0.24 \, \text{M}$

Find: The volume of acid solution in the form _____ \times 10^{-3} \, \text{mL}.

Since the acid is monobasic, the neutralization requires equal millimoles of acid and NaOH.

1. Calculate millimoles of NaOH:

$$\text{Millimoles of NaOH} = 0.24 \times 25 = 6 \, \text{mmol}$$

2. Therefore, millimoles of acid required = $6 \, \text{mmol}$.

3. Mass of acid required:

$$\text{Mass of acid} = 6 \times 24.2 = 145.2 \, \text{mg}$$

4. Volume of acid solution: Using

$$\text{Volume} = \frac{\text{mass}}{\text{density}}$$ $$V = \frac{145.2}{1.21 \times 10^3} = 0.12 \, \text{mL}$$

5. Convert to the required form:

$$V = 12 \times 10^{-3} \, \text{mL}$$

Therefore, the required volume is $12 \times 10^{-3} \, \text{mL}$. Hence, the answer is $12$.