Henry's Law constant of methyl bromide, CH3Br, is k = 0.159 mol/(L ∙
atm) at 25°C. What is the solubility of methyl bromide in water at 25°C and at
a partial pressure of 270. mm Hg?
Answers
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.
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.