99 Percentile Qs Bank for JEE MainChemistrySolutions
The vapour pressure of water at T (K) is 20 mm Hg. The following solutions are prepared at T (K) I. 6 g of urea (molecular weight = 60) is dissolved in 178.2 g of water. II. 0.01 mol of glucose is dissolved in 179.82 g of water. III. 5.3 g of N a 2 C O 3 (molecular weight = 106) is dissolved in 179.1 g of water. Identify the correct order in which the vapour pressure of solutions increases
Options
- AIII < I < II
- BII < III < I
- CI < II < III
- DI < III < II
Correct answer
A. III < I < II
Step-by-step solution
I. X B = n B n A + n B = 6 60 178.2 18 + 6 60 = 0.01 Δ P P 0 = X B = 0.01 II. X B = n B n A + n B = 0.01 179.82 18 + 0.01 = 0.001 Δ P P 0 = 0.001 III. X B = n B n A + n B = 5.3 106 179.1 18 + 5.3 106 = 0.005 Δ P P 0 = i X B = 3 × 0.005 = 0.015 So vapour pressure of solution will increase in the following sequence. III < I < II