Question

The Henry's law constant (k_h) for Ar in water at 20°C is 0.0015 mol/(L atm). How many grams of gas will dissolve in 2.78 L of H₂O in contact with pure Ar at 1.6 atm?

Answer 1

Answer:- 0.267 g

Explanation:-

Given:-

Henry's law constant for Ar gas

(K​​​​​​h) = 0.0015 mol/L.atm

Pressure of Ar gas = P = 1.6 atm

Volume of H​​​​​2O = 2.78 L

First we calculate the solubility of Ar gas in 1 L H​​​​​​2​O at 1.6 atm

We know that,

According to Henry's law,.    .....(1)

where,  

S = solubility of the gas

K​​​​​h = Henry's law constant of the gas

P = pressure of the gas

Therefore, from equation (1) we have,

  

Ie. 0.0024 mol of Ar gas dissolved in 1 L of H2O

Now,

We calculate the solubility of Ar gas in 2.78 L of H2O

We have,

1 L H2O = 0.0024 mol of Ar

Therefore,

2.78 L H2O =

= 0.00667 mol Ar

Now, we calculate mass of Ar dissolved

we have, Molar Mass of Ar = 39.95 g/mol

Formula :-

  

Therefore,

Therefore,

0.267 grams of Ar gas will dissolve in 2.78 L of H2O in contact with pure Ar at 1.6 atm.