Question

At 25.0°C the Henry's Law constant for methane CH4 gas in water is ×1.410−3/Matm.

Calculate the mass in grams of CH4 gas that can be dissolved in 75.mL of water at 25.0°C and a CH4 partial pressure of 0.68atm. Round your answer to 2 significant digits.

Answer 1

1.1 × 10⁻³ g

Step-by-step explanation:

Step 1: Given data

Henry's Law constant for methane (k): 1.4 × 10⁻³ M/atm

Volume of water (=volume of solution): 75 mL

Partial pressure of methane (P): 0.68 atm

Step 2: Calculate the concentration of methane in water (C)

We will use Henry's law.

`C = k * P = 1.4 * 10^(-3)M/atm * 0.68atm = 9.5 * 10^(-4)M`

Step 3: Calculate the moles of methane in 75 mL of water

`(9.5 * 10^(-4)mol)/(L) * 0.075 L = 7.1 * 10^(-5)mol`

Step 4: Calculate the mass corresponding to 7.1 × 10⁻⁵ mol of methane

The molar mass of methane is 16.04 g/mol.

`7.1 * 10^(-5)mol * (16.04g)/(mol) = 1.1 * 10^(-3) g`