Q33 / 97 · NTA Abhyas (JEE Main) 2020 · Chemistry · Solutions
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 (molecular weight = 106) is dissolved in 179.1 g of water.
Identify the correct order in which the vapour pressure of solutions increases
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 (molecular weight = 106) is dissolved in 179.1 g of water.
Identify the correct order in which the vapour pressure of solutions increases
- Correct answerAIII < I < II
- BII < III < I
- CI < II < III
- DI < III < II
Solution
I.
II.
III.
So vapour pressure of solution will increase in the following sequence.
III < I < II
II.
III.
So vapour pressure of solution will increase in the following sequence.
III < I < II
Hence, the correct answer is option: 'A'.