Question

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

(density of water = 10³ kg m^–3 at 298 K) This table implies that : 

(1) The pressure of a 55.5 molal solution of  γ is 1 bar 

(2) The pressure of a 55.5 molal solution of δ is 250 bar 

(3) Solubility of γ at 308 K is lower than at 298 K 

(4) α  has the highest solubility in water at a given pressure 

Answer 1

Correct option is (2) The pressure of a 55.5 molal solution of δ is 250 bar

(1) For γ

\((p)_{\gamma} = (K_H)_{\gamma} . (x) _{\gamma}\)

\(= 2 \times 10^{-5} \left [ \frac{55.5}{55.5 + \frac{1000}{18}}\right]\)

\(= 2 \times 10^{-5} \) K bar

\(= 2 \times 10^{-2}\) K bar

(2) For δ

\((p)_{\delta} = (K_H)_{\delta} . (x) _{\delta}\)

\(= 0.5 \left [ \frac{55.5}{55.5 + \frac{1000}{18}}\right]\)

\(= 0.5 \times 0.5\)

\(= 0.252\) K bar

= 250 bar

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

(4) From Henry's law p = K_H(x)

Higher the value of K_H, smaller will be the solubility of the gas, soy is more soluble.

Answer 2

(2) The pressure of a 55.5 molal solution of δ is 250 bar 

(1) P_\(\gamma\) = K_HX_Y

(3) On increasing temperature solubility of gases decreases 

(4) K_H \(\downarrow\) solubility ­\(\uparrow\) and lowest K_H is for \(\gamma\)