Question

1.

a.)Lauryl alcohol, C₁₂H₂₅OH, is prepared from coconut oil and used to make sodium lauryl sulfate, a synthetic detergent. What is the molality of lauryl alcohol in a solution of 17.1 g of lauryl alcohol dissolved in 185 g ethanol, C₂H₅OH?

b.). The carbohydrate digitoxose contains 48.64% carbon and 8.16% hydrogen; the remaining mass is oxygen. The addition of 18.0 g of this compound to 100. g of water gives a solution that has a freezing point of -2.2°C. What is the molecular formula of the compound? What is the molar mass of this compound to the nearest tenth of a gram?

c.) A gaseous mixture consists of 88.0 mole percent methane, CH₄, and 12.0 mole percent C₂H₆(ethane). Suppose water is saturated with the gas mixture at 20°C and 1.00 atm total pressure, and then the gas is expelled from the water by heating. What is the composition in mole fractions of the gas mixture that is expelled? The solubilities of CH₄and C₂H₆at 20C, and 1.00 atm are 0.023 g/L H₂O and 0.059 g/L H₂O, respectively.

Answer 1

Post one more question to get the answer to the last problem

1)

a) Molar mass of Lauryl Alcohol (C12H25OH) = 12 * 12 + 26 * 1 + 1 * 16 = 144 + 26 + 16 = 186 gm/mol

Number of moles of lauryl alcohol = 17.1/186 = 0.09193 moles

molality = number of moles of solute/per Kg of solvent

=> 0.09193/(185/1000)

=> 0.4969 m

b)

Depression in freezing point = i * Kf * m

2.2 = 1 * 1.86 * m

m = 1.1827

1.1827 = number of moles of solute/per kg of solvent

1.1827 = 10 * number of moles of solute

moles of solute = 0.11827 moles

Molar mass of compound = 18/0.11827 = 152.18 gm/mol

Mass of carbon in sample = 48.64/100 * 152.18 = 74.02 ( hence it contains 6 atoms of carbons)

Mass of hydrogen = 8.16/100 * 152.18 = 12.41 (12 atoms of carbon)

Mass of Oxygen = (100-48.64-8.16)/100 * 152.18 = 4 atoms of oxygen

Hence the compound is C6H12O4