At 10°c one volume of water dissolves 3.10 volumes of chlorine gas at 1.00 atm pressure. what is the henry's law constant of cl2 in water? (r = 0.0821 atm • l • mol-1• k-1)

Question

Trey Gardner

Chemistry

10 March, 23:33

At 10°c one volume of water dissolves 3.10 volumes of chlorine gas at 1.00 atm pressure. what is the henry's law constant of cl2 in water? (r = 0.0821 atm • l • mol-1• k-1)

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Terrell Bradley · Answer 1

Answer is: 0,133 mol / l· atm.

T (chlorine) = 10°C = 283K.

p (chlorine) = 1 atm.

V (chlorine) = 3,10 l.

R - gas constant, R = 0.0821 atm·l/mol·K.

Ideal gas law: p·V = n·R·T

n (chlorine) = p·V : R·T.

n (chlorine) = 1atm · 3,10l : 0,0821 atm·l/mol·K · 283K = 0,133mol.

Henry's law: c = p·k.

k - Henry's law constant.

c - solubility of a gas at a fixed temperature in a particular solvent.

c = 0,133 mol/l.

k = 0,133 mol/l : 1 atm = 0,133 mol / l· atm.