The Henry's Law constant (k) for carbon monoxide in water at 25°C is 9.71 x 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? A) 3.53 x 10-4 g D) 7.48 x 10-2 g B) 2.67 x10-3 g E) None of these choices is correct. C) 9.89 x 10-3 g

2.67 * 10-3 mol/L

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 proportionality factor is called Henry's law constant. (Wikipedia)

Given that; C =kP

C=concentration of a dissolved gas = ?

k=Henry's Law constant = 9.71 x 10-4 mol/(L·atm)

P=partial pressure of the gas = 2.75 atm

So,

C = 9.71 x 10-4 mol/(L·atm) * 2.75 atm

C = 2.67 * 10-3 mol/L

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm