Question
Calculate the solubility of certain gas in solvent with pressure 3 atm at 25^ C (Henry's law constant is 3.0 10^ -2 \, mol dm ^ -3 \, atm ^ -1 )
Calculate the solubility of certain gas in solvent with pressure 3 atm at 25^ C (Henry's law constant is 3.0 10^ -2 \, mol dm ^ -3 \, atm ^ -1 )
C. 0.09 M
According to Henry's law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. S = K_H P Given: K_H = 3.0 10^ -2 mol dm ^ -3 atm ^ -1 P = 3 atm Substituting the values: S = (3.0 10^ -2 mol dm ^ -3 atm ^ -1 ) 3 atm S = 9.0 10^ -2 mol dm ^ -3 Since 1 mol dm ^ -3 = 1 M , the solubility is 0.09 M . Answer: 0.09 M
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