SolutionMedium
Question
The elevation in boiling point, when 13.44 g of freshly prepared CuCl2 is added to one kg of water, is (Some useful data: Kb of water = 0.52 K − kg/mol and molecular weight of CuCl2 = 134.4)
Options
A.0.052
B.0.104
C.0.156
D.0.208
Solution
$\Delta T_{b\left( \text{theo} \right)} = K_{b}.m = 0.52 \times \frac{13.44}{134.4} = 0.052\text{ K}$
$i = 1 + \alpha(n - 1) = 1 + 1(3 - 1) = 3 $$$\therefore\Delta T_{b\left( \text{exp} \right)} = i.\Delta T_{b\left( \text{theo} \right)} = 3 \times 0.52 = 0.156\text{ K}$$
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