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Chemistry Question on Colligative Properties

The boiling point of water (100∘C)\left(100^{\circ} C \right) becomes 100.52∘C100.52^{\circ} C, if 33 grams of a non-volatile solute is dissolved in 200 ml200 \,ml of water. The moleular weight of solute is (if KbK _{ b } for water is =0.6 K/m=0.6\, K / m )

A

17.3 g.mol−117.3\, g.mol ^{-1} :

B

15.4 g.mol−115.4\, g.mol ^{-1} :

C

12.2 g.mol−112.2\, g.mol ^{-1} :

D

20.4 g.mol−120.4\, g.mol ^{-1} :

Answer

17.3 g.mol−117.3\, g.mol ^{-1} :

Explanation

Solution

As deration in boilding point is given by: ΔT=Kb×\Delta T=K_{b} \times molality Putting the various values, we get: 100.52−100.00=0.6×3/M200/1000100.52-100.00=0.6 \times \frac{3 / M}{200 / 1000} ⇒M=17.3 g mol−1\Rightarrow M=17.3\, g\, mol ^{-1}