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 mmHgI have been asked this question during an online exam.The above asked question is from Solubility Solutions in section Solutions of Chemistry – Class 12

At NTP, the solubility of natural gas in water is 0.8 mole of gas/kg o

1.	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 mmHgI 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» 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.

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