$\mathbf{M X}_{\mathbf{2}}$ dissociates into $\mathbf{M}^{\mathbf{2}+}$ and $\mathbf{X}^{-}$ ions in an aqueous solution, with a degree of dissociation $(\alpha)$ of 0.5. The ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of freezing point in the absence of ionic dissociation is

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Solutions

MX_{2}

dissociates into

M^{2 +}

and

X^{-}

ions in an aqueous solution, with a degree of dissociation

(α)

of 0.5. The ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of freezing point in the absence of ionic dissociation is

Answer: 2

Solution

The compound

MX_{2}

dissociates as:

MX_{2} \to M^{2 +} + 2 X^{-}

Degree of dissociation

α = 0.5

.

Initially: 1 mole of

MX_{2}

After dissociation:
•

MX_{2}

remaining:

1 - α = 0.5

•

M^{2 +}

formed:

α = 0.5

•

X^{-}

formed:

2 α = 1.0

Total particles =

(1 - α) + α + 2 α = 1 + 2 α

Van't Hoff factor

i = 1 + 2 α = 1 + 2 (0.5) = 2

Depression of freezing point:
• With dissociation:

Δ T_{f} = i \cdot K_{f} \cdot m

• Without dissociation (

i = 1

Δ T_{f}^{0} = K_{f} \cdot m

Ratio =

\frac{Δ T _{f}}{Δ T _{f}^{0}} = \frac{i \cdot K _{f} \cdot m}{K _{f} \cdot m} = i = 2

Answer: 2

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Integer answer

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JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Solutions

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