At 25 ° C, 13.0 mg of CO dissolves in 500 g of water when the partial pressure of CO is 0.968 atm. What is the Henry's law constant (in molal/atm) for CO at this temperature?

Question

Quizplus · Accepted Answer

Henry's law constant, kH, is given by the equation:

kH = (mole fraction of gas in solution) / (partial pressure of gas)

First, we need to find the mole fraction of CO in solution. This can be calculated using the formula:

mole fraction = moles of solute / moles of solute + moles of solvent

We are given that 13.0 mg of CO dissolves in 500 g of water. The molar mass of CO is 28.01 g/mol, so the number of moles of CO is:

moles of CO = 13.0 mg / (28.01 g/mol) = 4.642 × 10^-4 mol

The number of moles of water is:

moles of water = 500 g / (18.02 g/mol) = 27.74 mol

Therefore, the mole fraction of CO in solution is:

mole fraction of CO = 4.642 × 10^-4 mol / (4.642 × 10^-4 mol + 27.74 mol) = 1.67 × 10^-5

Next, we need to convert the partial pressure of CO into units of atm. We are given that the partial pressure is 0.968 atm.

Finally, we can use the above equation to calculate the Henry's law constant:

kH = (mole fraction of CO in solution) / (partial pressure of CO)
   = (1.67 × 10^-5) / (0.968 atm)
   = 1.73 × 10^-5 molal/atm

This value matches answer choice C.