Question

At 100°C the vapour pressure of a solution of 6.5g of a solute in 100 g water is 732 mm. If $K_b$ = 0.52, the boiling point of this solution will be
(a) 101°C
(b) 100°C
(c) 102°C
(d) 103°C

Answer 1

Dear khushi,

We know that relative lowering of vapour pressure is given by the formula

P0A− PAP0A = Xsolute

P0A of pure water = 760 mm Hg

PA of the solution = 732 mm Hg

This implies Xsolute = 760−732760 = 0.037

Now as we have got the mole fraction of the solute , we can find the no. of moles of solute present

0.037 = nsolutensolute + 10018

This gives , nsolute = 0.213

Hence we have 0.213 moles of the solute present . Now we know the formula for elevation in boiling point.

ΔTb = kb * m = 0.52 * 0.2131001000 = 1.1076

Hence ΔTb = 1.1076

Therefore the boiling point will be = (100 + 1.1076) = 101.1076o C

Hence the boiling point of the solution will be 101.1076o C.

I hope you understood .

So option (a)is correct

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