Question

What is the solubility of methylacetylene (in units of grams per liter) in water at 25 °C, when the C3H4 gas over the solution has a partial pressure of 0.301 atm? kH for C3H4 at 25 °C is 9.23×10-2 mol/L·atm.

Answer 1

The solubility of methylacetylene is 0,11 g L⁻¹

Step-by-step explanation:

Henry's law is a gas law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid.

The formula is:

C = kH P

Where C is solubility of the gas (In mol/L)

kH is Henry constant (9,23x10⁻² mol L⁻¹ atm⁻¹)

An P is partial pressure (0,301 atm)

Solving, C = 2,78x10⁻³ mol L⁻¹. In grams per liter:

2,78x10⁻³ mol L⁻¹ₓ
`(40 g)/(mol)` = 0,11 g L⁻¹

I hope it helps!