The vapour pressure of water at T(K) is 20 mm Hg. The following solutions are prepared at T(K):
I. 6g of urea (mol.wt.=60) is dissolved in 178.2 g of water.
II. 0.01 mole of glucose is dissolved in 179.82 g of water.
III. 5.3 g of `Na_2CO_3` (mol.wt. =106) is dissloved in 179.1 g of water.
Identify the correct order in which the vapour pressures of solutions increase:

Options

A. III, I, II
B. II, III, I
C. I, II, III
D. I, III, II

Answer

A

Explanation

I. `x_(B)= (n_(B))/(n_(A) + n_(B)) = (6//60)/(178.2//18+6//60)= 0.01`
`(Delta P)/(p_(0))= x_(B)= 0.01`
II. `x_(B)= (n_(B))/(n_(A) + n_(B)) = (0.01)/((179.82)/(18) + 0.01)`
`(Delta 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`
`(Delta p)/(p_(0)) = ix_(B)= 3 xx 0.005= 0.015`
`therefore` Vapour pressure of solutions will increase in the following sequence. `(III) lt (I) lt (II)`