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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Dalton’s Law of Partial Pressure/ Henry"s law

(b) 15.03 cmCH4(NTP) is dissolved in 500 cm3 water.

Thus, we have Volume v dissolved in 1 cm3 water = 15.30500×1=0.03006cm3

Amount of gas dissolved = PVRT=1×0.0300682.06×273

                                    =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

=1×10-3×16300×1=5.33×10-5gm

From Henry's law, pressure required to dissolve the above amount = mk=5.33×10-52.144×10-5=2.486 atm.

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