Question

Imagine that 27.0 g of C2H2(g) dissolves in 1.00 L of liquid acetone at 1.00 atm pressure. If the partial pressure of C2H2(g) is increased to 12.0 atm, what is its solubility in acetone?

Answer 1

Answer: The solubility of $C_2H_2(g)$ at 12.0 atm is $324g/1.00L$

Explanation:

We are given:

Mass of $C_2H_2(g)$ = 27.0 grams

Volume of liquid acetone = 1.00 L

Solubility of $C_2H_2(g)$ 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:

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

Or,

$\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}$

where,

$C_1\text{ and }p_1$ are the initial concentration and partial pressure of $C_2H_2(g)$

$C_2\text{ and }p_2$ are the final concentration and partial pressure of $C_2H_2(g)$

We are given:

$C_1=27.0g/1.00L\\p_1=1.00atm\\C_2=?\\p_2=12.0atm$

Putting values in above equation, we get:

$\frac{27.0g/1.00L}{C_2}=\frac{1.00atm}{12.0atm}\\\\C_2=\frac{27.0g/1.00L\times 12.0atm}{1.00atm}=324g/1.00L$

Hence, the solubility of $C_2H_2(g)$ at 12.0 atm is $324g/1.00L$