At 298 K, 500cm3 H2O dissolved 15.30 cm3 CH4(STP) under a partial pressure of methane of one atm. If Henery's law holds, what pressure is required to cause 0.001 mole methane to dissolve in 300cm3 water ?
(a) 0.286 atm
(b) 2.486 atm
(c) 1.286 atm
(d) 3.111 atm
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(b) 15.03 cm3 CH4(NTP) is dissolved in 500 cm3 water.
Thus, we have Volume v dissolved in 1 cm3 water = cm3
Amount of gas dissolved =
=1.34 X 10-6mol
Mass of gas disolved, m=1.34 X 10-6 X 16 = 2.144 X 10-5gm
Now, according to Henry's law, we have m=Kp (K- Henry constant. K=m/p.
Substituting the values of m and p, we get K=2.144 X 105gm atm-1
Now for 1 X 10-3mole of gas in 300cm-3 of water we have Mass of gas dissolved in 1 cm3 water
From Henry's law, pressure required to dissolve the above amount =