Question
Which of the following $$0.10$$ $$M$$ aqueous solution will have the lowest freezing point?
A.
$$A{l_2}{\left( {S{O_4}} \right)_3}$$
B.
$${C_5}{H_{10}}{O_5}$$
C.
$$Kl$$
D.
$${C_{12}}{H_{22}}{O_{11}}$$
Answer :
$$A{l_2}{\left( {S{O_4}} \right)_3}$$
Solution :
Depression in freezing point $$ \propto $$ number of particles.
In colligative properties ions behave like particles.
$$A{l_2}{\left( {S{O_4}} \right)_3}$$ provides five ions on ionisation as
$$\eqalign{
& A{l_2}{\left( {S{O_4}} \right)_3} \to 2\,A{l^{3 + }} + 3SO_4^{2 - } \cr
& KI \rightleftharpoons {K^ + } + {I^ - } \cr} $$
while $$KI$$ provides two ions and $${C_5}{H_{10}}{O_5}$$ and $${C_{12}}{H_{22}}{O_{11}}$$ are not ionised, so they have single particle
Hence, lowest freezing point is possible for $$A{l_2}{\left( {S{O_4}} \right)_3}.$$