A very diluted saturated solution of a sparingly soluble salt `X_(3)Y_(4)` has a vapour pressure of 20 mm Hg temperature T,while pure water exerts a pressure of 20.0126 mm Hg at the same temperature . Calculate molality (m)at temperature T :

Answer

`5.4 xx10^(-13)`

Explanation

`P_(s) = 20 P^(@) = 20.0126`
`(P^(@)-P_(s))/(P^(@)) = (0.0126)/(20) = (n)/(n+N) =(n)/(N)`
`("mole of solute")/("moles of" H_(2)O) = 0.0063`
1 mole `H_(2)O = 18g = 18 mL`
`18 mL` solution `= 0.0063` mole
`1L` solution `= (0.00063)/(18) xx 1000 = 0.35` mole/L
Let solubility of salt `A_(3)B_(4)` is s then
`7s = 0.035`
`s = 0.005` mole/L
`k_(sp) = 3^(3).4^(4) (s)^(7) = 27 xx 256 xx (0.005)^(7)`
`k_(sp) = 5.4 xx 10^(-13)`