Question

the vapour pressure of 2.1 %solution of a non-electrolyte in water at 100C is 755mmHg.Calculate the molar mass of the 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 (Ans)

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Answer 2

Answer:

The molar mass of non-electrolyte solute is equal 58.69gmol⁻¹.

Explanation:

We know that vapor pressure of water  at 100°C, $P_{o}=760mmHg$

Given the vapor pressure of the solution $P=755mmHg$

Because solution has 2.1% non-electrolyte

Mass of non-electrolyte solute, $W_{2}=2.1grams$

Rest is mass of water,  $W_{1}=100-2.1=97.9g$

$M_{1}=18gmol^{-1}$ molar mass of water and $M_{2}$ is molar mass of solute

Now        $\frac{P_{o}-P}{P_{o}} =\frac{W_{2}*M_{1}}{W_{1}*M_{2}}$

∴            $M_{2}=\frac{W_{2}*M_{1}}{W_{1}}(\frac{P_{o}}{P_{o}-P} )$

             $M_{2}=\frac{(700)(2.1)(18)}{(760-755)*97.9}$

            $M_{2}=58.69gmol^{-1}$

Therefore the molar mass of non-electrolyte solute is 58.69gmol⁻¹.