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)

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 $> KCl > BaCl _{2}> AlCl _{3}$