Question

1) What is the solubility of cyclopropane (in units of grams per liter) in water at 25 °C, when the C₃H₆ gas over the solution has a partial pressure of 0.217 atm? k_H for C₃H₆ at 25 °C is 1.20×10^-2 mol/L·atm.

___g/L

2)What is the solubility of neon (in units of grams per liter) in water at 25 °C, when the Ne gas over the solution has a partial pressure of 273 mm Hg? k_Hfor Ne at 25 °C is 4.51×10^-4 mol/L·atm.

___g/L

Answer 1

To calculate the solubility of a gas in a liquid using Henry's law, we can use the following formula:

Solubility (in g/L) = (kH * P_gas) / Molar mass of the gas

Where: kH is the Henry's law constant in units of mol/L·atm, P_gas is the partial pressure of the gas in atm, and Molar mass of the gas is in g/mol.

Let's calculate the solubility of cyclopropane and neon in water at 25 °C:

1. Solubility of cyclopropane (C3H6) in water at 25 °C: Partial pressure of C3H6 gas (P_gas) = 0.217 atm Henry's law constant for C3H6 at 25 °C (kH) = 1.20 × 10^-2 mol/L·atm Molar mass of C3H6 = 3 × 12.01 g/mol + 6 × 1.01 g/mol ≈ 42.08 g/mol

Solubility (in g/L) = (1.20 × 10^-2 mol/L·atm * 0.217 atm) / 42.08 g/mol Solubility ≈ 6.17 × 10^-5 g/L

The solubility of cyclopropane in water at 25 °C is approximately 6.17 × 10^-5 g/L.

2. Solubility of neon (Ne) in water at 25 °C: Partial pressure of Ne gas (P_gas) = 273 mm Hg Convert mm Hg to atm: 1 atm ≈ 760 mm Hg, so 273 mm Hg ≈ 0.359 atm Henry's law constant for Ne at 25 °C (kH) = 4.51 × 10^-4 mol/L·atm Molar mass of Ne = 20.18 g/mol

Solubility (in g/L) = (4.51 × 10^-4 mol/L·atm * 0.359 atm) / 20.18 g/mol Solubility ≈ 8.02 × 10^-6 g/L

The solubility of neon in water at 25 °C is approximately 8.02 × 10^-6 g/L.