for water is \[1.86\,\,K\,\,kg\,mo{{l}^{-1}},\] the lowering in freezing point of the solution is:
A) -1.12 K
B) 0.56 K
C) 1.12 K
D) -0.56K
Correct Answer: C
Solution :
\[\underset{\begin{smallmatrix} weak\,acid \\ 1 \\ 1-\alpha \end{smallmatrix}}{\mathop{HX}}\,\rightleftharpoons \,\underset{\begin{smallmatrix} 0 \\ \alpha \end{smallmatrix}}{\mathop{{{H}^{+}}}}\,+{{\underset{\begin{smallmatrix} 0 \\ \alpha \,\, \end{smallmatrix}}{\mathop{{{X}^{-}}}}\,}_{(at\ equilibrium)}}\] \[\alpha =20%\,dissociation\] i.e., \[\alpha \] = 0.2 \[i=1-\alpha +\alpha +\alpha \] \[=1+\alpha =1+0.2=1.2\] \[\Delta {{T}_{f}}=i\times {{K}_{f}}\times m\] \[=1.2\times 1.86\,K\,kg\,mo{{l}^{-1}}\,\times 0.5\] = 1.12 K
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