You visited us 1 times! Enjoying our articles? Unlock Full Access!

Depression in Freezing Point

The freezing ...

Question

The freezing point depression of $10^{- 3}$ molal aqueous solution of a compound $K_{x} [F e (C N)_{6}]$ is $7.44 \times 10^{- 3}$ K. Find $x$ .
Given : $K_{f}$ of $H_{2} O = 1.86$ K mol $^{- 1}$ kg

Open in App

Solution

The expression for the depression in the freezing point is $Δ T_{f} = i \times k_{f} \times m$ .

Substitute values in the above expression.

$7.44 \times 10^{- 3} K = i \times 1.86 k m o l^{- 1} k g \times 10^{- 3} m$

$i = 4$

But $i = \frac{n (o b s e r v e d)}{n (t h e o r e t i c a l)}$

Hence, $n (o b s e r v e d) = 4 \times n (t h e o r e t i c a l)$

Thus, the dissociation of 1 molecule of $K_{x} [F e (C N)_{6}]$ gives 4 ions. This is possible only if the value of x is 3.

$K_{3} [F e (C N)_{6}] ⇌ 3 K^{+} + [F e (C N)_{6}]^{3 -}$

Suggest Corrections

Similar questions

0.001 m K_{x} [F e (C N)_{6}]

7.44 \times 10^{- 3} K

(K_{f})_{w a t e r} = 1.86 K k g m o l^{- 1}

K_{x} [F e (C N)_{6}]

1.00 \times 10^{- 3}

K_{x} [F e (C N)_{6}]

7.35 \times 10^{- 3}

m o l^{- 1},

{- 0.385}^{o} C

K_{f} = 3.85

K

k g

{m o l}^{- 1}

K_{b} = 0.712

K

k g

{m o l}^{- 1}

_{2}

_{f}

_{2}

^{- 1}

_{b}

_{2}

^{- 1}]

0.1 M H A (a q)

- {0.2046}^{\circ} C

p H

K_{f} o f H_{2} O = 1.86 K k g m o l^{- 1}

Join BYJU'S Learning Program