Answer :
Answer: The solubility of [tex]C_2H_2(g)[/tex] at 12.0 atm is [tex]324g/1.00L[/tex]
Explanation:
We are given:
Mass of [tex]C_2H_2(g)[/tex] = 27.0 grams
Volume of liquid acetone = 1.00 L
Solubility of [tex]C_2H_2(g)[/tex] in liquid acetone at 1.00 atm = 27.0 g / 1.00 L
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 [tex]C_2H_2(g)[/tex]
[tex]C_2\text{ and }p_2[/tex] are the final concentration and partial pressure of [tex]C_2H_2(g)[/tex]
We are given:
[tex]C_1=27.0g/1.00L\\p_1=1.00atm\\C_2=?\\p_2=12.0atm[/tex]
Putting values in above equation, we get:
[tex]\frac{27.0g/1.00L}{C_2}=\frac{1.00atm}{12.0atm}\\\\C_2=\frac{27.0g/1.00L\times 12.0atm}{1.00atm}=324g/1.00L[/tex]
Hence, the solubility of [tex]C_2H_2(g)[/tex] at 12.0 atm is [tex]324g/1.00L[/tex]