Skip to content

Osmosis And Reverse Osmosis

Easychemistry

The osmotic pressure of a 0.02 M solution of a non-electrolyte at 300 K is closest to which value? (Take R = 0.082 L·atm·K⁻¹·mol⁻¹)

Select the correct option:

🔒 Solution Hidden from View

Submit your answer to unlock the detailed step-by-step solution.

About This Question

Subject
chemistry
Chapter
solutions
Topic
osmosis and reverse osmosis
Difficulty
Easy
Year
2025
Tags
osmotic pressurevan't Hoff equationdilute solutionscolligative propertiesconcentration

Solution

Correct Answer:

0.492 atm

Osmotic pressure is calculated using the van't Hoff equation: π = CRT, where C is the molar concentration, R is the gas constant, and T is the absolute temperature. For a non-electrolyte, the van't Hoff factor i = 1. Given: C = 0.02 mol/L, R = 0.082 L·atm·K⁻¹·mol⁻¹, T = 300 K. Therefore, π = 0.02 × 0.082 × 300 = 0.492 atm. Option 0.248 atm corresponds to using T = 150 K (half the given temperature), an arithmetic slip. Option 0.984 atm results from using C = 0.04 mol/L (doubling the concentration). Option 0.124 atm would require C = 0.005 mol/L, one-fourth of the given concentration. The van't Hoff equation for osmotic pressure has the same mathematical form as the ideal gas law, which is its physical origin — the partial pressure of the solute behaving like an ideal gas. Plausibility check: a small concentration (0.02 M) gives a small osmotic pressure (< 1 atm), which is physically sensible. This is a direct NCERT example from the Solutions chapter and is a staple JEE Main numerical question.

This easy difficulty chemistry question is from the chapter solutions, covering the topic of osmosis and reverse osmosis. It appeared in the 2025 exam.

Looking for more practice? Explore all chemistry questions or browse solutions questions on RankGuru.