Methane (CH4, molar mass = 16.0 g/mol) has a Henry's Law constant (kH) of 9.88 × 10–2 mol/(L·atm) when dissolved in benzene at 25 °C. How many grams of CH4will dissolve in 3.00 L of benzene if the partial pressure of CH4 is 1.48 atm?

The mass of methane that will dissolve is 0.7008 grams

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

where,

= Henry's constant =

= molar solubility of methane gas = ?

= partial pressure of methane gas = 1.48 atm

Putting values in above equation, we get:

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

Given mass of methane = ?

Molar mass of methane = 16.0 g/mol

Molarity of solution = 0.0146 M

Volume of solution = 3.00 L

Putting values in above equation, we get:

Hence, the mass of methane that will dissolve is 0.7008 grams

h let me think