The vapour pressure of a saturated solution of sparingly soluble salt `("XCl"_(3))` was 17.20mm Hg at `27^(@)"C"`. If the vapour pressure of pure `"H"_(2)"O"` is 17.25 mm Hg at 300K, what is the solubility of sparingly soluble salt `"XCl"_(3)" in mole"//"Litre".`

Options

A. `4.04xx10^(-2)`
B. `8.08xx10^(-2)`
C. `2.032xx10^(-2)`
D. `4.04xx10^(-2)`

Answer

A

Explanation

Let solubility of `"XCl"_(3)="S mol lit"^(-1)`
For `"XCl"_(3),"n"=4,alpha=1`
`therefore"i"=4`
As S is very small
`therefore` Molality=Molarity
`("P"^(0)-"P"_("solution"))/("P"^(0))xx"i"xx("n"_("solute"))/("n"_("solvent")`
`(17.25-17.20)/17.25=(4"S")/55.55`
`"S"=4.04xx10^(-2)"mol lit"^(-1)`