Henry\'s law may be expressed in different ways and with different concentration

Question

Henry's law may be expressed in different ways and with different concentration units, resulting in different values for the Henry's law constants. If mole fraction is used as the concentration unit, one algebraic statement of the law is:

Pgas = kHXgas   where k is the Henry's law constant in units of pressure, usually atm.

At 25°C, some water is added to a sample of gaseous butane (C4H10) at 3.54 atm pressure in a closed vessel and the vessel is shaken until as much butane as possible dissolves. Then 6.81 kg of the solution is removed and boiled to expel the butane, yielding a volume of 0.681 L of C4H10(g) at 0°C and 1.00 atm. Determine the Henry’s law constant for butane in water based on this experiment.

_______________________atm

Explanation / Answer

Assuming ideal behaviour,

P V = n R T

1.00 * 0.681 = n * 0.0821 * 273.15

n = 0.0304 mol

Mass of butane = 0.0304 * 58 = 1.76 g.

Therefore,

Mass of Solvent i.e water = mass of solution - mass of solute = 6810 - 1.76 = 6808.24 g.

Moles of water = 6808.24 / 18 = 378.2 mol

Mole fraction of butane = moles of butane / (total moles of solution) = 0.0304 / (0.0304 + 378.2) = 0.0000804

THerefore,

Pgas = kH * Xgas

3.54 = kH * 0.0000804

kH = 44044. atm.

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