Question

The vapour pressure of a solvent at 293 K is 100 mm Hg. Then the vapour pressure of a solution containing 1 mole of a strong electrolyte $(AB_2)$ in 99 moles of the solvent at 293 K is (assume complete dissociation of solute)

• A. 103 mm Hg
• B. 99 mm Hg
• C. 97 mm Hg
• D. 101 mm Hg

Answer 1

Answer: (C)

97 mm Hg

Answer 2

Let us consider the problem:
$\frac{\left( P _{ A }{ }^{0}- P ^{0}\right)}{ P _{A}{ }^{0}}= i \frac{\left( n _{ B }\right)}{\left( n _{ A }+ n _{ B }\right)}$
Since,
$\frac{\left(100-P_{A}\right)}{100}=3\left[\frac{1}{(99+1)}\right]$
Since,
$100- P _{ A }=3$
Hence the solution is $P _{ A }=97\, mm$ of $Hg$.