Question

At NTP, the solubility of natural gas in water is 0.8 mole of gas/kg of water. What is the Henry’s law constant for natural gas?

(a) 8 kN/m^2

(b) 7.90 x 10^-3 Pa

(c) 71.36 bar

(d) 105 mmHg

I have been asked this question during an online exam.

The above asked question is from Solubility Solutions in section Solutions of Chemistry – Class 12

Answer 1

Right choice is (c) 71.36 bar

Easiest explanation: Given,

Solubility of natural gas, S = 0.8 mole of gas/kg of water = 0.8 molal

Hence, moles of natural gas in mixture, nNG = 0.8 mole

At NTP, pressure, PNG = 1.01325 bar

Number of moles of water in 1000 g, nw = mass/molar mass

nw = 1000g / (18g/mole) = 55.56 mole

Mole fraction of natural gas in the mixture, XNG = nNG/ (nNG + nw)

XNG = 0.8/(0.8 + 55.56) = 0.0142

Using Henry’s law PNG = KH x XNG

KH = PNG/XNG = 1.01325 bar/0.0142 = 71.36 bar.