Question

The henry's law constant (k) for carbon monoxide in water at 25°c is 9.71 × 10-4 mol/(l·atm). how many grams of co will dissolve in 1.00 l of water if the partial pressure of co is 2.75 atm?

Answer 1

Using Henry's Law, 0.0748 grams of carbon monoxide (CO) will dissolve in 1.00 L of water when the partial pressure of CO is 2.75 atm, with the provided Henry's Law constant of 9.71 × 10^-4 mol/(L·atm).

Step-by-step explanation:

The student has asked how many grams of CO will dissolve in 1.00 L of water if the partial pressure of CO is 2.75 atm, given that the Henry's law constant (k) for carbon monoxide in water at 25°C is 9.71 × 10-4 mol/(L·atm). To answer this question, we first use Henry's Law, which states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. The formula for Henry's Law is:S = kP

Where S is the solubility of the gas (in mol/L), k is the Henry's Law constant, and P is the partial pressure of the gas. Plugging in the values, we get:

S = (9.71 × 10-4 mol/(L·atm)) × (2.75 atm) = 2.67 × 10-3 mol/L

Now that we have the solubility, we can calculate the mass of CO that will dissolve by converting moles to grams using the molecular weight of CO (28.01 g/mol):

Mass of CO = S × molecular weight of CO = (2.67 × 10-3 mol/L) × (28.01 g/mol) = 0.0748 g

Therefore, 0.0748 grams of CO will dissolve in 1.00 L of water at a partial pressure of 2.75 atm.

Answer 2

In this question, you are given the Henry constant (9.71 × 10-4 mol/l atm), the volume(1 l) and the pressure(2.75 atm). The molecular mass of CO should be 28g/mol.
Then the amount of carbon monoxide in the water in grams should be: 9.71 × 10-4 mol/l atm * 1l * 2.75atm * 28g/mol= 747.67 *10^-4= 0.074767 grams