Question

Correct order of freezing point of $1\,M$ solution of sucrose, $KCl,\,BaCl_{2}$ and $AlCl_{3}$ is

• A. Sucrose $>KCl>BaCl_{2}>AlCl_{3}$
• B. $AlCl_{3}>BaCl_{2}>KCl >$ Sucrose
• C. $BaCl_{2}>KCl>AlCl_{3}>$ Sucrose
• D. $KCl>BaCl_{2}>AlCl_{3}>$ Sucrose

Answer 1

Answer: (A)

Sucrose $>KCl>BaCl_{2}>AlCl_{3}$

Answer 2

Depression in freezing point is a colligative property, it means that as the number of particles (ions) of solute increases, freezing point of the solution decreases. Sucrose $\longrightarrow$ not ionised $KCl \rightleftharpoons \underbrace{ K ^{+}+ Cl ^{-}}_{2 \text{ions} }$ $BaCl _{2} \rightleftharpoons \underbrace{ Ba ^{2+}+2 Cl ^{-}}_{3 \text { ions }}$ $AlCl _{3} \rightleftharpoons \underbrace{ Al ^{3+}+3 Cl ^{-}}_{4 \text { ions }}$ Hence, the correct order of freezing point is Sucrose $> KCl > BaCl _{2}> AlCl _{3}$