Question
The elevation in boiling point of a solution of $$13.44 g$$ of $$CuC{l_2}$$ in 1 kg of water using the following information will be ( Molecular weight of $$CuC{l_2} = 134.4\,{\text{and}}\,{K_b} = 0.52\,mola{l^{ - 1}}$$ )
A.
0.16
B.
0.05
C.
0.1
D.
0.2
Answer :
0.16
Solution :
TIPS/Formulae :
$$\eqalign{
& (i)\,i = \frac{{{\text{No}}{\text{. of particles after ionisation }}}}{{{\text{No}}{\text{. of particles before ionisation}}}} \cr
& (ii)\,\Delta {T_b} = i \times {K_b} \times m \cr} $$
\[\underset{\begin{smallmatrix}
1 \\
\left( 1-\alpha \right)
\end{smallmatrix}}{\mathop{CuC{{l}_{2}}}}\,\to \underset{\begin{smallmatrix}
0 \\
\alpha
\end{smallmatrix}}{\mathop{C{{u}^{2+}}}}\,+\underset{\begin{smallmatrix}
0 \\
2\alpha
\end{smallmatrix}}{\mathop{2C{{l}^{-}}}}\,\]
$$\eqalign{
& i = \frac{{1 + 2\alpha }}{1}\,\,i = 1 + 2\alpha \cr
& {\text{Assuming }}100\% {\text{ ionization}}\,\,\,\,\,\,\,\,\,So,\,i = 1 + 2 = 3 \cr
& \Delta {T_b} = 3 \times 0.52 \times 0.1 = 0.156 \approx 0.16\,\,\,\left[ {m = \frac{{13.44}}{{134.4}} = 0.1} \right] \cr} $$