Answer:
1.1 × 10⁻³ g
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.
[tex]C = k \times P = 1.4 \times 10^{-3}M/atm \times 0.68atm = 9.5 \times 10^{-4}M[/tex]
Step 3: Calculate the moles of methane in 75 mL of water
[tex]\frac{9.5 \times 10^{-4}mol}{L} \times 0.075 L = 7.1 \times 10^{-5}mol[/tex]
Step 4: Calculate the mass corresponding to 7.1 × 10⁻⁵ mol of methane
The molar mass of methane is 16.04 g/mol.
[tex]7.1 \times 10^{-5}mol \times \frac{16.04g}{mol} = 1.1 \times 10^{-3} g[/tex]
