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 )
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.
