Exercise 12.98 A gas has a Henry's law constant of 0.136 M/atm Part A How much...

Exercise 12.98 A gas has a Henry's law constant of 0.136 M/atm Part A How much water would be needed to completely dissolve 1.53 L of the gas at a pressure of 745 torr and a temperature of 23 ∘C?

Solutions

Expert Solution

henry's law...
Pgas = kH x c
or
c = kH x Pgas

Both of those equations are common henry's law expressions and you have to look at the units of kH to know what equation you're dealing with. kH either has units of Latm / mole = atm / M.. or... mole / Latm... M / atm...you have the second...so.. which equation do you use?

look at the first...P = kH x c.... ---> P / c = kH.. atm / M... you don't have that...
now look at the second.. c = kH x P.. c / P = kH.. M / atm... you do have that..

so for this problem...

c = kH x Pgas

where...
Pgas = partial pressure of gas above liquid
kH = henry's law constant
c = concentration of gas in the liquid state.

************
c = 745 torr x (1atm / 760 torr) x ((0.136 moles / L) / atm) = 0.133 moles / L liquid

and from PV = nRT
n = PV / RT..
n = (745 torr x 1atm/760 torr) x (1.53L) / [(0.08206 Latm/moleK) x (296K)] = 0.0617 moles gas

then finally...
0.0617 moles gas x (1L liquid / 0.133 moles) = 0.463 L of water

queen_honey_blossom answered 1 year ago

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