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Question: An aqueous solution of 2% non volatile solute exerts a pressure of 1.004 bar at the normal boiling p...

An aqueous solution of 2% non volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent what is the molecular mass of the solute?

A

23.4gmol−123.4gmol^{- 1}

B

41.35gmol−141.35gmol^{- 1}

C

10gmol−110gmol^{- 1}

D

20.8gmol−120.8gmol^{- 1}

Answer

41.35gmol−141.35gmol^{- 1}

Explanation

Solution

Vapour pressure of pure water at boiling point

= 1atm=1.013bar1atm = 1.013bar

Vapour pressure of solution (ps)=1.004bar(p_{s}) = 1.004bar

Let mass of solution = 100 g

Mass of solute = 2g

Mass of solvent = 100−2=98g100 - 2 = 98g

Pº−PsPº=n1n1+n2=W2/M2W1/M1\frac{Pº - P_{s}}{Pº} = \frac{n_{1}}{n_{1} + n_{2}} = \frac{W_{2}/M_{2}}{W_{1}/M_{1}}

1.013−1.0041.013=2M2×1898\frac{1.013 - 1.004}{1.013} = \frac{2}{M_{2}} \times \frac{18}{98}

Or M2=2×1898×1.0130.009=41.35gmol−1M_{2} = \frac{2 \times 18}{98} \times \frac{1.013}{0.009} = 41.35gmol^{- 1}