Question
What will be the freezing point of a $$0.5\,m\,KCl$$ solution ? The molal freezing point constant of water is $${1.86^ \circ }C\,{m^{ - 1}}.$$
A.
$$ - {1.86^ \circ }C$$
B.
$$ - {0.372^ \circ }C$$
C.
$$ - {3.2^ \circ }C$$
D.
$${0^ \circ }C$$
Answer :
$$ - {1.86^ \circ }C$$
Solution :
$$\eqalign{
& \Delta {T_f} = i{K_f} \times m \cr
& \,\,\,\,\,\,\,\,\,\,\, = 2 \times 1.86 \times 0.5 \cr
& \,\,\,\,\,\,\,\,\,\,\, = {1.86^ \circ }C \cr
& {T_f} = T_f^ \circ - \Delta {T_f} \cr
& \,\,\,\,\,\,\, = 0 - 1.86 \cr
& \,\,\,\,\,\,\, = - {1.86^ \circ }C \cr} $$