An aqueous solution is prepared by dissolving 0 . 1 mole of an ionic salt in 1 . 8   kg of water at 35 ° C . The salt remains 90 % dissociated in the solution. The vapour pressure of the solution is 59 . 724   mm of Hg . Vapor pressure of water at 35 ° C is 60 . 000   mm…

Question

An aqueous solution is prepared by dissolving 0 . 1 mole of an ionic salt in 1 . 8 &#160; kg of water at 35 &#176; C . The salt remains 90 % dissociated in the solution. The vapour pressure of the solution is 59 . 724 &#160; mm of Hg . Vapor pressure of water at 35 &#176; C is 60 . 000 &#160; mm of Hg . The number of ions present per formula unit of the ionic salt is _____.

Aicharya · Accepted Answer

Number of ions present per formula unit of ionic salt $= x$
Van 't Hoff factor
$(i) = 1 + (n - 1) α = 1 + (x - 1) \times 0.9 = 0.1 + 0.9 x$
(Assuming $90 %$ dissociation)
$∴$ Relative lowering in vapour pressure $= i \times$ Mole fraction of solute
$\Rightarrow \frac{60 - 59.724}{60} = \frac{i \times 0.1}{\frac{1800}{18} + 0.1}$
$\Rightarrow 0.0046 = \frac{i \times 0.1}{100 + 0.1}$
$0.0046 \approx \frac{(0.9 x + 0.1) \times 0.1}{100}$
$\Rightarrow 0.9 x + 0.1 = 4.6$
$\Rightarrow x = \frac{4.5}{0.9} = 5$
$x = 5$

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An aqueous solution is prepared by dissolving 0 . 1 mole of an ionic salt in 1 . 8   kg of water at 35…

Solution

An aqueous solution is prepared by dissolving 0 . 1 mole of an ionic salt in 1 . 8 &#160; kg of water at 35…

Solution

An aqueous solution is prepared by dissolving 0 . 1 mole of an ionic salt in 1 . 8 kg of water at 35…