The degree of dissociation of Ca NO 32 in a dilute aqueous solution containing 7 g of salt in 100 g of water at 100- C is 70 - The vapour pressure of solution is-A- 746-27 mm of HgB- 745-27 mm of HgC- 744-27 mm of HgD- 743-27 mm of Hg

The correct option is A 746.27 mm of HgWe have, Ca(NO_3)_2↔ Ca^2+ + 2NO^ 3 Before dissociation 1 nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp ...

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Standard XII

Chemistry

Abnormal Colligative Properties

The degree of...

Question

The degree of dissociation of $C a (N O_{3})_{2}$ in a dilute aqueous solution containing 7 g of salt in 100 g of water at $100^{\circ} C$ is $70 %$ . The vapour pressure of solution is:

746.27 mm of Hg

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745.27 mm of Hg

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744.27 mm of Hg

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743.27 mm of Hg

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Solution

The correct option is A 746.27 mm of Hg
We have, $C a (N O_{3})_{2} \leftrightarrow C a^{2 +} + 2 N O^{- 3}$
Before dissociation 1 0 0
After dissociation 1 $- α$ $α$ 2 $α$
Therefore, total moles at equilibrium = $1 + 2 α$
$= 1 + 2 \times 0.7 [A s, α = 0.7] = 2.4$
For $C a (N O_{3})_{2}$ : $\frac{m_{N}}{m_{e x p}} = 1 + 2 α$
Therefore, $m_{e x p} = \frac{m_{N}}{(1 + 2 α)} = \frac{164}{2.4} = 68.33$
Also at $100^{\circ} C, P^{0} H_{2} O = 760 m m, w = 7 g, W = 100 g$
And $\frac{(P^{0} - P_{s})}{P_{s}} = \frac{(7 \times 18)}{(68.33 \times 100)} = 0.0184 \frac{P^{0}}{P_{s}} - 1 = 0.0184 ∴ P_{s} = \frac{760}{1.0184} = 746.27 m m o f H g$

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The degree of dissociation of Ca(NO3)2 in a dilute aqueous solution containing 7 g of salt in 100 g of water at 100∘C is 70%. The vapour pressure of solution is:

The degree of dissociation of $C a (N O_{3})_{2}$ in a dilute aqueous solution containing 7 g of salt in 100 g of water at $100^{\circ} C$ is $70 %$ . The vapour pressure of solution is: