Q19 / 150 · MHT CET 2026 (18 April Shift 2) · Chemistry
Calculate the molar mass of nonelectrolyte solute when 6 gram of it is dissolved in 1 $\text{dm}^3$ water has osmotic pressure $2.4\ \text{atm}$ at 300 K $(R = 0.0821\ \text{atm dm}^3\ \text{K}^{-1}\ \text{mol}^{-1})$
- A$45.0\ \text{g mol}^{-1}$
- B$74.12\ \text{g mol}^{-1}$
- Correct answerC$61.58\ \text{g mol}^{-1}$
- D$90.28\ \text{g mol}^{-1}$
Solution
The osmotic pressure is given by the formula:
$\pi = C R T = \dfrac{W R T}{M V}$
Rearranging the formula to solve for the molar mass $M$:
$M = \dfrac{W R T}{\pi V}$
Substituting the given values $W = 6\ \text{g}$, $R = 0.0821\ \text{atm dm}^3\ \text{K}^{-1}\ \text{mol}^{-1}$, $T = 300\ \text{K}$, $\pi = 2.4\ \text{atm}$, and $V = 1\ \text{dm}^3$:
$M = \dfrac{6 \times 0.0821 \times 300}{2.4 \times 1}$
$M = \dfrac{147.78}{2.4}$
$M = 61.575\ \text{g mol}^{-1}$
Rounding to two decimal places, the molar mass is $61.58\ \text{g mol}^{-1}$.
Answer: $61.58\ \text{g mol}^{-1}$
$\pi = C R T = \dfrac{W R T}{M V}$
Rearranging the formula to solve for the molar mass $M$:
$M = \dfrac{W R T}{\pi V}$
Substituting the given values $W = 6\ \text{g}$, $R = 0.0821\ \text{atm dm}^3\ \text{K}^{-1}\ \text{mol}^{-1}$, $T = 300\ \text{K}$, $\pi = 2.4\ \text{atm}$, and $V = 1\ \text{dm}^3$:
$M = \dfrac{6 \times 0.0821 \times 300}{2.4 \times 1}$
$M = \dfrac{147.78}{2.4}$
$M = 61.575\ \text{g mol}^{-1}$
Rounding to two decimal places, the molar mass is $61.58\ \text{g mol}^{-1}$.
Answer: $61.58\ \text{g mol}^{-1}$
Hence, the correct answer is option: 'C'.