0.01 M solution of KCl and BaCl 2 are prepared in water. The freezing point of KCl found to be -4° C. What will be the freezing point for BaCl 2 solution assuming that both KCl and BaCl2 are completely ionised in solutions?

Question

0.01 M solution of KCl and BaCl 2 are prepared in water. The freezing point of KCl found to be -4° C. What will be the freezing point for BaCl 2 solution assuming that both KCl and BaCl2 are completely ionised in solutions? (A)  -3°C  (B)  -4°C  (C)  -5°C  (D)  -6°C

Tardigrade Admin · Accepted Answer

KClK ++ Cl -(2 ions) BaCl 2 Ba 2++2 Cl -(3 ions) ∵ 2 ions shows depression in freezing point =-4° C. ∴ 3 ions shows depression in freezing point =(-4/2) × 3=-6° C

Q. $0.01 M$ solution of $K Cl$ and $B a C l_{2}$ are prepared in water. The freezing point of $K Cl$ found to be $- 4^{\circ} C$ . What will be the freezing point for $B a C l_{2}$ solution assuming that both $K Cl$ and $B a C l_{2}$ are completely ionised in solutions?

$- 3^{\circ} C$

$- 4^{\circ} C$

$- 5^{\circ} C$

$- 6^{\circ} C$

$K Cl K^{+} + C l^{-}$ (2 ions)
$B a C l_{2} B a^{2 +} + 2 C l^{-}$ (3 ions)
$∵ 2$ ions shows depression in freezing point $= - 4^{\circ} C .$
$∴ 3$ ions shows depression in freezing point
$= \frac{- 4}{2} \times 3 = - 6^{\circ} C$

Q. 0.01M solution of KCl and BaCl2​ are prepared in water. The freezing point of KCl found to be −4∘C. What will be the freezing point for BaCl2​ solution assuming that both KCl and BaCl2​ are completely ionised in solutions?

−3∘C

−4∘C

−5∘C

−6∘C