Question

The vapour pressure of 2.1 % solution of a non electrolyte in water at hundred degree celsius is 755 mm hg calculate the molar mass of solute

Answer 1

Hi friend

Here is your answer

The vapour pressure of the pure water should be given to solve the problem. Let it be 760 mm Hg (p0).

Let the mass of water be 100 g (W1). So mass of non-electrolyte =2.1 g(W2).

Molar mass of water (M1)=18 g. Molar mass of non-electrolyte(M2)=?

vapour pressure of the solution = 755 mm Hg.

So by relative lowering of vapour pressure

(p0-p)/p0 =i X

as it is non-electrolyte so i=1

(760-755)/760 = n2/(n1+n2) ≈ n2/n1 [ solution is dilute (2.1%) so n2 « n1]

or, 5/760 =(W2 x M1)/(M2xW1) = (2.1 x 18)/(M2 x 100)

or, M2 = 57.456 g mol-1

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