Henry's law constant for solubility of methane In benzene is ` 4.2xx 10^(-5) mm` Hg at a particular constant temperature At this temperature calculate the solubility of methane at
(i) 750 mm Hg (ii) 840 mm Hg

Explanation

` (K_H)_(benzene ) = 4.2 xx 10 ^(-5) mm Hg`
Solubility of methane =?
` P = 750 mm Hg `
According to Henry,s Law
` P =K_H . X _( "in solution ") `
750 mm Hg = 4.2 ` xx 10^(- 5) mm Hg X_( " in solution ") `
` rArr " " X_ ( " in solution ") " " = ( 75)/( 4.2 xx 10^(-5) ) `
i.e solubility = ` 178.5 xx 10 ^(-5) `
similary at P =` 840 mm Hg `
Solubility =` ( 840)/( 4.2 xx 10 ^(-5) ) = 200 xx 10^(-5) `