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Chemistry Question on Solutions

An aqueous solution of 2%2\% non-volatile solute exerts a pressure of 1.0041.004 bar at the normal boiling point of the solvent. What is the molecular mass of the solute?

A

23.4 g mol−123.4\, g\, mol^{-1}

B

41.35 g mol−141.35\, g\, mol^{-1}

C

10 g mol−110\, g\, mol^{-1}

D

20.8 g mol−120.8\, g\, mol^{-1}

Answer

41.35 g mol−141.35\, g\, mol^{-1}

Explanation

Solution

Vapour pressure of pure water at boiling point =1 atm=1.013 bar=1\, atm = 1.013 \,bar Vapour pressure of solution (ps)=1.004 bar(ps) = 1.004\, bar Let mass of solution =100 g= 100\, g Mass of solute =2 g= 2 \,g Mass of solvent =100−2=98 g= 100 - 2 = 98\, g p∘−psp∘=n2n1+n2\frac{p^{\circ}-p_{s}}{p^{\circ}} = \frac{n_{2}}{n_{1}+n_{2}} =n2n1=W2/M2W1/M1(∵ n2<<<n1)=\frac{n_{2}}{n_{1}}=\frac{W_{2}/M_{2}}{W_{1}/M_{1}} \left(\because\,n_2 <<< n_{1}\right) 1.013−1.0041.013=2M2×1898\frac{1.013-1.004}{1.013} = \frac{2}{M_{2}} \times \frac{18}{98} or M2=2×1898×1.0130.009M_{2 } = \frac{2 \times 18}{98} \times \frac{1.013}{0.009} =41.35 g mol−1= 41.35\,g\,mol^{-1}