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Solutions question

2014 · Shift 1 · Q4

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JEE Advanced
Chemistry
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2014 · Shift 1 · Q4

Solutions question

2014 · Shift 1 · Q4

JEE AdvancedChemistrySolutionsNumerical+3 / −1

MX_2

dissociates in

M^{2+}

and

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

Numerical answer

View written solutionFree

Correct answer: 2

Write the dissociation equation

For the salt $MX_2$ : $MX_2 \rightarrow M^{2+} + 2X^-$

So, if dissociation were complete, one formula unit of $MX_2$ would give 3 particles in solution.

Use degree of dissociation $\alpha = 0.5$

Let initially there be $1$ mole of $MX_2$ .

Undissociated moles of $MX_2 = 1-\alpha = 1-0.5 = 0.5$
Dissociated moles of $MX_2 = \alpha = 0.5$

From $0.5$ mole dissociated:

$M^{2+}$ formed $= 0.5$ mole
$X^-$ formed $= 2 \times 0.5 = 1.0$ mole

Total particles present after dissociation

Total solute particles $= 0.5 + 0.5 + 1.0 = 2.0$

Thus, the van't Hoff factor is $i = 2$

Alternatively, for $MX_2$ giving 3 ions: $i = 1 + (3-1)\alpha = 1 + 2(0.5) = 2$

Relate freezing point depression to van't Hoff factor

Depression in freezing point is: $\Delta T_f = iK_f m$

In absence of dissociation: $i=1$
Observed value: $i=2$

Therefore, $\frac{\text{observed } \Delta T_f}{\text{value in absence of dissociation}} = \frac{2K_f m}{1K_f m} = 2$

Final answer

The required ratio is: $\boxed{2}$

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