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 = $\frac{15.30}{500}×1=0.03006$cm3

Amount of gas dissolved = $\frac{\mathrm{PV}}{\mathrm{RT}}=\frac{1×0.03006}{82.06×273}$

=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

$=\frac{1×{10}^{-3}×16}{300}×1=5.33×{10}^{-5}\mathrm{gm}$

From Henry's law, pressure required to dissolve the above amount =

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