Question

What is the concentration of N2 in a fresh water stream in equilibrium with air at 298 K and 1 atmosphere? Given the value of KH for N2 = 0.00060 mole/kgbar.

(a) 0.0474 g/kg

(b) 0.0005 g/kg

(c) 1316.7 g/kg

(d) 13.3 g/kg

The question was posed to me in exam.

My question is taken from Solubility Solutions topic in chapter Solutions of Chemistry – Class 12

Answer 1

Right option is (d) 13.3 g/kg

The best I can explain: Given,

KH = 0.00060 mole/kgbar

PN2 = mole fraction of N2 x Pair (from Dalton’s law)

Air consists of 79 mole% N2 and 21 mole% O2.

PN2 = 0.79 x 1 bar= 0.79 bar

Henry’s law –PN2x KH= solubility of N2

0.79 bar x 0.00060 (mole/kgbar) = solubility of N2

Solubility of N2 = 4.74 x 10^-4 moles of N2/kg water

Converting moles of N2 to kg of N2 :

Solubility of N2 = 4.74 x 10^-4 mole x 28 kg/mole = 0.0133 kg N2/kg water = 13.3 g N2/kg water.