Question

The solubility of h2s in water at stp is 0.195mol kg_1 calculate Henry's law constant

Answer 1

Answer:

0.987/0.0035=282 atm

Explanation:

Given Molarity =0.195m

Molarity =

mass of solvent (in kg)

no. of moles of solute

Let us consider mass of water =1kg

∴ moles of solute = molarity ×mass of solvent

=0.195×1=0.195mol

Molar mass of H

2

O=2×1+16=18

No. of moles of water =

molarmass

mass

=

18

1000

=55.56mol (∵ 1kg−1000g)

mole fraction of H

2

S=

(0.195+55.56)

0.195

=0.0035

At STP,

P=0.987 atm

According to Henry's law

P=K

H

×x

K

H

=

x

p

=0.987/0.0035=282 atm