Question

Henry’s law is important in environmental chemistry, where it predicts the distribution of pollutants between water and the atmosphere. The hydrocarbon cyclopropane (C3H6) emitted in wastewater streams, for example, can pass into the air, where it is degraded by processes induced by light from the sun. The Henry’s law constant for cyclopropane in water at 25 °C is 4273 atm, when the following form of the law is used: Pcyclopropane = kcyclopropane Xcyclopropane Calculate the partial pressure of cyclopropane vapor in equilibrium with a solution of 1.84 g of cyclopropane per 1010 L of water. How many cyclopropane molecules are present in each cubic centimeter of vapor?

Answer 1

Step-by-step explanation:

Henry's law states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid

number of moles of cyclopropane = 1.84 g / 42.08 g/mol = 0.044 mol

number of mole of water = 1010000 g / 18.02 g/mol = 56048.83 mol

mole fraction of cyclopropane = mole of cyclopropane / ( mole of water + mole of cyclopropane) = 0.044 mol / 56048.87 = 7.85 × 10⁻⁷

P cyclopropane = k henry × mole fraction of cyclopropane = 4273 atm × 7.85 × 10⁻⁷ = 0.00335 atm

b) number of cyclopropane molecules present in each cubic centimeter = ( 0.044 mol / 1010) × 6.022 × 10 ²³ / 1000 cm³ = 2.623 × 10 ¹⁶ molecules in cm³