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?

Answer :

Answer: 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:

[tex]C_{CH_4}=K_H\times p_{CH_4}[/tex]

where,

[tex]K_H[/tex] = Henry's constant = [tex]9.88\times 10^{-2}mol/L.atm[/tex]

[tex]C_{CH_4}[/tex] = molar solubility of methane gas = ?

[tex]p_{CH_4}[/tex] = partial pressure of methane gas = 1.48 atm

Putting values in above equation, we get:

[tex]C_{CH_4}=9.88\times 10^{-2}mol/L.atm\times 1.48atm\\\\C_{CH_4}=0.0146M[/tex]

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

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]

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:

[tex]0.0146=\frac{\text{Mass of methane}}{16.0g/mol\times 3.00}\\\\\text{Mass of methane}=(0.0146\times 16\times 3)=0.7008g[/tex]

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

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