A 0.04 M solution has an osmotic pressure of 1.476 atm at 300 K. If the true molar mass is 150 g mol⁻¹, what will be the observed molar mass?
Answer and explanation
Correct answer: 100 g mol⁻¹
From π = iCRT, the van’t Hoff factor is i = 1.476/(0.04 × 0.082 × 300) = 1.50. Since i is greater than 1, dissociation has increased the number of particles, causing the observed molar mass to be smaller than the true value. The relation is Mobserved = Mtrue/i = 150/1.50 = 100 g mol⁻¹. Therefore, option C is correct.
Frequently asked questions
What is the correct answer to this question?
100 g mol⁻¹
Why is this the correct answer?
From π = iCRT, the van’t Hoff factor is i = 1.476/(0.04 × 0.082 × 300) = 1.50. Since i is greater than 1, dissociation has increased the number of particles, causing the observed molar mass to be smaller than the true value. The relation is Mobserved = Mtrue/i = 150/1.50 = 100 g mol⁻¹. Therefore, option C is correct.
Which subject and chapter does this question cover?
This is a Class 12 Chemistry question. Chapter: Chapter 01: Solutions. Topic: 6: Molar Mass Determination.
