Henry's constant (in kbar) for four gases α , β , γ and δ in water at 298   K…

Henry's constant (in kbar) for four gases α,β,γ  and δ in water at 298 K is given below :

αβγδKH5022×10-50.5

(density of water =103 k gm-3 at 298 K ) This table implies that :

  1. α has the highest solubility in water at a given pressure
  2. solubility of γ at 308K is lower than at 298K
  3. The pressure of a 55.5 molal solution of δ is 250 bar
  4. The pressure of 55.5 molal solution of γ is 1 bar.

Solution

(1) From Henery's law P=kHX

Higher the value of kH smaller will be solubility so γ is more soluble.

(2) Though solubility of gas will decrease with increase in temperature but this conclusion can not be drawn from the given table.

(3)For

δPδ=K.Xδ

=0.5×103×55.5555.55+100018=250bar.

(4)For γ

pγ=KHγ ×xγ

=2×10-255.555.5+100018=10-2bar

Asked in: JEE Main 2020 (03 Sep Shift 1)

Practice more Solutions questions on Aicharya