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crimeas [40]
3 years ago
13

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?
Chemistry
1 answer:
vladimir1956 [14]3 years ago
3 0

<u>Answer:</u> 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

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