Question

Problem 2.1 : The solubility of N, gas in water at 25 °C and 1 bar is 6.85 x 104 mol L'. Calculate (a) Henry's law constant (b) molarity of N, gas dissolved in water under atmospheric conditions when partial pressure of N, in atmosphere is 0.75 bar.​

Answer 1

Explanation:

Since in this problem ‘k’ is not given so we have to find out.

Applying Henry’s equation,S = k P

k= S/P = 6.8 X 10-4 mol L-1 = 6.8 X 10-4 mol L-1

1

So the solubility of N2 at 0.78 atm:

S= (6.8 X 10-4 mol L-1 ) (0.78 atm) = 5.3 x10-4 M

Answer 2

Given: concentration = $6.85*10^{-4} mol/L$ at 25 °C and 1 bar pressure.

To find: (i) henry's law constant.

             (ii) molarity at 0.75 bar.

Step-by-step explanation:

Step 1 of 2

According to Henry's Law $C=kP$.

$6.85*10^{-4} mol/L=k*1bar\\k=6.85*10^{-4} mol/Lbar$

Step 2 of 2

Molarity at 0.75bar can be calculated by Henry's law.

$C=kP\\C=6.85*10^{-4} mol/Lbar*0.75bar\\C=5.1375*10^{-4}mol/L$

HENRY"S CONSTANT FOR N IS $6.85*10^{-4} mol/Lbar$.

THE CONCENTRATION OF N AT 0.75BAR IS $5.1375*10^{-4}mol/L$.