At 25c, 13.0 mg of co dissolves in 500g of water when the partial pressure of co is 0.968 bar. what is the henry's law constant for co at this temperature

K = 9.59x10⁻⁴ m/bar

Explanation:

To do this, we need to write the Henry's law which is the following:

C = P*K  (1)

K is the constant of the henry's law

C is the concentration of the solution

P is the pressure.

We have data for pressure which is 0.968 bar, and we need to calculate the concentration.

Concentration of any solution is calculated with the expression of molarity, which is:

C = n/V  (2)     m = n/kg water

The volume of this solution, will be the volume of water. Assuming density of water = 1 g/mL we can say that 500 g of water = 500 mL of water or 0.5 kg of water. But this is a solvent, so, we will calculate the molality instead of molarity

The moles are calculated with:

n = m/MM  (3)

The molar mass of CO is:

MM CO = 12 + 16 = 28 g/mol

Now, let's use expression (3), (2) and finally the (1) to get the constant:

n = (13/1000) / 28 = 4.64x10⁻⁴ moles

Using expression (2), we will calculate the molality:

m = 4.64x10⁻⁴ / 0.5 = 9.28x10⁻⁴ m

Finally, from (1) we solve for K:

K = C/P

K = 9.28x10⁻⁴ / 0.968

K = 9.59x10⁻⁴ m/bar

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2