Question

Henry’s law constant for a gas CH3
Br is 0.159 moldm-3 atm at 250⁰c What is the solubility of CH3
Br in water at 25 ⁰c and a partial pressure of 0.164 atm?

Answer 1

Answer:

Explanation:

Given :- Henery's law constant of CHBr

K = 0.159 mol/(L.atm)

Temperature = 25°C

Partial pressure (P) = 270 am.Hg

To Find:- Solubility (S) of CHBr = ?

Solution:-

- The ability of a substance to dissolve in the solution is called solubility.

- In a liquid Solubility of gas is directly proportional to partial pressure of gas at constant temperature.

- The relation of Solubility with Henry law constant is

Solubility (Concentration) = Henry's law constant (mol/L.atm) × partial pressure (atm)

- Given, Partial pressure of CHBr

= 270 am.Hg

- As we know that,

1 atm = 760 mm.Hg

So, Partial pressure of CHBr will be,

= 270/760 atm

- By putting the value of partial pressure in the above formula we get,

Solubility (Concentration)

= 0.159 × 270/760

= 0.0564 or 5.64 × 10^(-2) mol/L

- Hence, The Solubility of CHBr is

5.64 × 10^(-2) mol/L.

Answer 2

Explanation:

P1/T1=P2/T2

0.164/(25+273)=P2/(250+273)

P2=0.288atm

S=KP2

S=0.159x0.288

S= 0.046moldm-³