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 Na2CO3 (molecular weight = 106) is dissolved in 179.1 g of water.
Identify the correct order in which the vapour pressure of solutions increases

  • Correct answer
    A
    III < I < II
  • B
    II < III < I
  • C
    I < II < III
  • D
    I < III < II

Solution

I. XB=nBnA+nB
=660178.218+660=0.01
ΔPP0=XB=0.01
II. XB=nBnA+nB
=0.01179.8218+0.01=0.001
ΔPP0=0.001
III. XB=nBnA+nB
=5.3106179.118+5.3106=0.005
ΔPP0=iXB
=3×0.005=0.015
So vapour pressure of solution will increase in the following sequence.
III < I < II

Hence, the correct answer is option: 'A'.

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