JEE Chemistry 2023 Question with Solution

Given: Molarity of NaCl solution is $3 \, \text{M}$, density is $1.0 \, \text{g mL}^{-1}$, and molar mass of NaCl is

$$23 + 35.5 = 58.5 \, \text{g mol}^{-1}$$

Find: Molality in the form $\_\_\_\_\_ \times 10^{-2} \, \text{m}$.

For $1 \, \text{L}$ of solution, mass of solution is

$$1.0 \, \text{g mL}^{-1} \times 1000 \, \text{mL} = 1000 \, \text{g}$$

Since the solution is $3 \, \text{M}$, moles of NaCl in $1 \, \text{L}$ are

$$3 \, \text{mol}$$

Mass of solute is

$$3 \times 58.5 = 175.5 \, \text{g}$$

Mass of solvent is

$$1000 - 175.5 = 824.5 \, \text{g} = 0.8245 \, \text{kg}$$

Now, molality is

$$m = \frac{3}{0.8245} = 3.64 \, \text{m}$$

Writing this as $\times 10^{-2}$,

$$3.64 = 364 \times 10^{-2}$$

Therefore, the required nearest integer is $364$.

Given: $M = 3$, density $d = 1.0 \, \text{g mL}^{-1}$, molar mass of solute $= 58.5 \, \text{g mol}^{-1}$.

A direct relation between molarity and molality is

$$m = \frac{1000M}{1000d - M \times \text{molar mass of solute}}$$

Substituting the given values,

$$m = \frac{1000 \times 3}{1000 \times 1 - 3 \times 58.5}$$ $$m = \frac{3000}{824.5} = 3.64$$

Hence,

$$m = 364 \times 10^{-2}$$

So, the correct numerical answer is $364$.