JEE Chemistry 2024 Question with Solution

Given: Mass of NaOH = $84 \, \text{g}$, molar mass of NaOH = $40 \, \text{g/mol}$, molarity = $3 \, \text{M}$.

Find: The value of $x$ in $x \times 10^{-1} \, \text{dm}^3$.

First calculate the number of moles of NaOH:

$$n_{\text{NaOH}} = \frac{84}{40} = 2.1$$

Now use the molarity relation:

$$M = \frac{n}{V}$$

So,

$$V = \frac{n}{M} = \frac{2.1}{3} = 0.7 \, \text{L}$$

Since $1 \, \text{L} = 1 \, \text{dm}^3$, the volume is $0.7 \, \text{dm}^3$.

Expressing this in the required form:

$$0.7 = 7 \times 10^{-1}$$

Therefore, the volume is $7 \times 10^{-1} \, \text{dm}^3$, so the answer is $7$.

Given: Mass of NaOH = $84 \, \text{g}$, formula weight = $40 \, \text{g/mol}$, required molarity = $3 \, \text{mol/L}$.

Find: The numerical value $x$.

From the mass and molar mass,

$$n_{\text{NaOH}} = \frac{\text{mass}}{\text{molar mass}} = \frac{84 \, \text{g}}{40 \, \text{g/mol}} = 2.1 \, \text{mol}$$

Using molarity,

$$M = \frac{n}{V_{\text{sol}}}$$

Rearranging,

$$V_{\text{sol}} = \frac{n_{\text{NaOH}}}{M} = \frac{2.1}{3} = 0.7 \, \text{L}$$

In scientific notation,

$$0.7 \, \text{L} = 7 \times 10^{-1} \, \text{L}$$

Also, $\text{L}$ and $\text{dm}^3$ are numerically equal.

Therefore, $x = 7$. The correct answer is 7.