Question

At the Henry's Law constant for carbon dioxide gas in water is . Calculate the mass in grams of gas that can be dissolved in of water at and a partial pressure of . Round your answer to significant digits.

Answer 1

The question is incomplete, here is the complete question:

At 25°C Henry's Law constant for carbon dioxide gas in water is 0.031 M/atm . Calculate the mass in grams of gas that can be dissolved in 425. mL of water at 25°C and at a partial pressure of 2.92 atm. Round your answer to 2 significant digits.

Answer: The mass of carbon dioxide that can be dissolved is 1.7 grams

Explanation:

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

$C_{CO_2}=K_H\times p_{CO_2}$

where,

$K_H$ = Henry's constant = $0.031M/atm$

$C_{CO_2}$ = molar solubility of carbon dioxide gas

$p_{CO_2}$ = partial pressure of carbon dioxide gas = 2.92 atm

Putting values in above equation, we get:

$C_{CO_2}=0.031M/atm\times 2.92 atm\\\\C_{CO_2}=0.0905M$

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

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of carbon dioxide = ? g

Molar mass of carbon dioxide = 44 g/mol

Molarity of solution = $0.0905mol/L$

Volume of solution = 425 mL

Putting values in above equation, we get:

$0.0905mol/L=\frac{\text{Mass of carbon dioxide}\times 1000}{44g/mol\times 425}\\\\\text{Mass of solute}=\frac{44\times 425\times 0.0905}{1000}=1.7g$

Hence, the mass of carbon dioxide that can be dissolved is 1.7 grams