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?

The solubility of at 12.0 atm is

Explanation:

We are given:

Mass of = 27.0 grams

Volume of liquid acetone = 1.00 L

Solubility of 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:

Or,

where,

are the initial concentration and partial pressure of

are the final concentration and partial pressure of

We are given:

Putting values in above equation, we get:

Hence, the solubility of at 12.0 atm is

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explanation:

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