Calculate Henry's law constant for `H_(2)S.` whose solubility in water at STP is assumed to be 0.195 m.

Answer

282

Explanation

(a) Calculation of mole fraction of `H _(2)S` 0.195 m means that 0.195 mole of `H _(2)S` dissolves in 1000 g of water.
Number of moles of water in 1000 g,
`( n _(H _(2)) O )= ( (1000 g ))/( ( 18 g mol ^(-1))) = 55.55 ` mol
Mole fraction `H _(2) S = ( n _(H _(2)S))/( n _(H _(2 )S) + n _(H _(2 ) ^(0)))`
`= (( 0. 195 mol))/( ( 0. 195 + 55.55 ) mol) = (( 0. 195 mol ))/((55.745mol )) = 0. 0035`
(b) Calculation of Henry.s law constant
According to Henry.s law
`x _( H _(2 )S) = ( "Partial pressure of " H _(2) S)/( K _(H) "for" H _(2)S) ` at STP
`K _(H)` for `H _(2) S = ("Partial pressrue of " H _(2)S)/( x _(H _(2)S))`
`= (( 0. 9 87 "bar" ))/( ( 0. 00 35)) = 282 ` bar