Respuesta :
Answer: The solubility of carbon dioxide at 5.50 atm is [tex]0.886g/100mL[/tex]
Explanation:
To calculate the molar solubility, we use the equation given by Henry's law, which is:
[tex]C_{A}=K_H\times p_{A}[/tex]
Or,
[tex]\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}[/tex]
where,
[tex]C_1\text{ and }p_1[/tex] are the initial concentration and partial pressure of carbon dioxide
[tex]C_2\text{ and }p_2[/tex] are the final concentration and partial pressure of carbon dioxide
We are given:
[tex]C_1=0.161g/100mL\\p_1=1.00atm\\C_2=?\\p_2=5.50atm[/tex]
Putting values in above equation, we get:
[tex]\frac{0.161g/100mL}{C_2}=\frac{1.00atm}{5.50atm}\\\\C_2=\frac{0.161g/100mL\times 5.50atm}{1.00atm}=0.886g/100mL[/tex]
Hence, the solubility of carbon dioxide at 5.50 atm is [tex]0.886g/100mL[/tex]
