Question

The partial pressure of CO2 in air at sea level is 0.239 torr. If the solubility of CO2 at this partial pressure and 25 °C is 0.034 M, calculate kH. What is the molarity of CO2 at Pco 7.60 torr. A solution contains a mixture of pentane (CsH12) and hexane (CsH14) at room temperature. The solution has a vapor pressure of 258 torr. Pure pentane has a vapor pressure of 425 torr and pure hexane has a vapor pressure of 151 torr at room temperature. Assuming ideal behavior, what is the mole fraction of pentane and the mole fraction of hexane

Answer 1

Using the Henry's law p proportionalto s solubility P = k_H middot s K_H = P/s = 0.239/0.034 partially pressure of CO_2 Then again using the above formula s = molarity = P/K_H = 7.6/0.2239/0.034 = 1.08M Pentane (C_5H_12) = P_1 degrees = 425 torr Hexane (C_6H_14) = P_2 degrees = 151 torr using Raoult's law P_total = P_1 degrees x_1 + P_2 degrees x_2, x_1, x_2 rightarrow mole fraction 285 = 425x_1 + 151x_2 x_1 + x_2 = 1 x_2 = 1 - x_1 = 425 x_1 + 151(1 - x_1) = 425x_1 - 151x_1 + 151 274x_1 = 134 x_1 = 134/274 = 0.489 = 0.49 Pentane x_2 = 1 - 0.49 = 0.51 for Hexane