JEE Chemistry 2023 Question with Solution

Given: HCl contains $73 \, \text{g L}^{-1}$ and $0.69 \, \text{g}$ of metallic sodium reacts with water.

Find: The volume of HCl required to completely neutralize the NaOH formed.

First, calculate moles of sodium:

$$\text{Moles of Na} = \frac{0.69}{23} = 3 \times 10^{-2}$$

The reaction is:

$$\text{Na} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \frac{1}{2} \text{H}_2$$

So, moles of NaOH produced = $3 \times 10^{-2}$.

Number of equivalents of NaOH = number of equivalents of HCl.

Mass concentration of HCl = $73 \, \text{g/L}$. Therefore, normality is:

$$\text{Normality} = \frac{73}{36.5} = 2 \, \text{N}$$

Using

$$\text{N}_1 V_1 = \text{N}_2 V_2$$

we get

$$2 \times V = 3 \times 10^{-2}$$

Hence,

$$V = 15 \, \text{mL}$$

Therefore, the required volume is $15 \, \text{mL}$, so the numerical answer is 15.

Given: Sodium reacts with water to form NaOH, which is then neutralized by HCl.

Find: The nearest integer value of the volume of acid required.

Since NaOH is formed in a $1:1$ ratio from sodium,

$$\text{moles of NaOH} = \text{moles of Na} = 3 \times 10^{-2}$$

For monoprotic HCl and monobasic NaOH, equivalent factor is $1$ for both. Hence,

$$\text{equivalents of NaOH} = 3 \times 10^{-2}$$

and this must equal equivalents of HCl used.

Now convert the acid concentration to normality:

$$\text{Normality of HCl} = \frac{73 \, \text{g/L}}{36.5 \, \text{g/equiv}} = 2 \, \text{equiv/L}$$

So volume required is

$$V = \frac{3 \times 10^{-2}}{2} \; \text{L} = 1.5 \times 10^{-2} \; \text{L} = 15 \, \text{mL}$$

Therefore, the correct numerical value is 15.