Question
The difference between the boiling point and freezing point of an aqueous solution containing sucrose ( molecular $$wt = 342\,g\,mol{e^{ - 1}}$$ ) in $$100\,g$$ of water is $${105^ \circ }C.$$ If $${K_f}$$ and $${K_b}$$ of water are $$1.86$$ and $$0.51\,K\,kg\,mo{l^{ - 1}}$$ respectively, the weight of sucrose in the solution is about
A.
34.2$$\,g$$
B.
342$$\,g$$
C.
7.2$$\,g$$
D.
72$$\,g$$
Answer :
72$$\,g$$
Solution :
$$\eqalign{
& \left( {100 + \Delta {T_b}} \right) - \left( {0 - \Delta {T_f}} \right) = 105 \cr
& \Delta {T_b} + \Delta {T_f} = 5 \cr
& m\left( {{K_b} + {K_f}} \right) = 5 \cr
& m = \frac{5}{{2.37}}\,\,i.e.,\frac{5}{{2.37}}moles\,{\text{in}}\,1000g\,{\text{water}} \cr
& \left( {{\text{or}}} \right)\frac{5}{{2.37 \times 10}}moles\,{\text{in}}\,100g\,{\text{water}} \cr
& \therefore \,\,Wt.\,{\text{of sucrose}} = \frac{5}{{2.37 \times 10}} \times 342 = 72\,g \cr} $$