Question

$0.01\, M$ solution of $KCl$ and $BaCl _{2}$ are prepared in water. The freezing point of $KCl$ found to be $-4^{\circ} C$. What will be the freezing point for $BaCl _{2}$ solution assuming that both $KCl$ and $BaCl_{2}$ are completely ionised in solutions?

• A. $ -3^{\circ}C $
• B. $ -4^{\circ}C $
• C. $ -5^{\circ}C $
• D. $ -6^{\circ}C $

Answer 1

Answer: (D)

$KClK ^{+}+ Cl ^{-}$(2 ions)
$BaCl _{2} Ba ^{2+}+2 Cl ^{-}$(3 ions)
$\because 2$ ions shows depression in freezing point $=-4^{\circ} C.$
$\therefore 3$ ions shows depression in freezing point
$=\frac{-4}{2} \times 3=-6^{\circ} C$