How much water would be needed to completely dissolve 1.64 of the gas at a pressure of 750 torr and a temperature of 21 C? Henrys law constant is .145 M/atm.

1.64 WHAT?

To determine the amount of water needed to completely dissolve a gas, we can use Henry's law equation:

C = k * P

Where:
C is the concentration of the dissolved gas in water (in Molarity or moles per liter)
k is Henry's law constant (in M/atm)
P is the partial pressure of the gas (in atmospheres)

In this case, we are given:
Henry's law constant (k) = 0.145 M/atm
Partial pressure (P) = 750 torr = 750/760 atm (since 1 atm = 760 torr)

Now let's substitute the given values into the equation and calculate the concentration (C):

C = 0.145 M/atm * (750/760 atm)
C ≈ 0.143 M

This means that the concentration of the gas in water would be approximately 0.143 M.

To calculate the amount of water needed to completely dissolve the gas, we need to know the volume of the solution we want to prepare. Let's assume we want to prepare 1 liter of the solution.

The amount of solute (here, the gas) is given by:
Amount = Concentration * Volume

Amount = 0.143 M * 1 L
Amount ≈ 0.143 moles

Therefore, to completely dissolve 1.64 moles of the gas, approximately 0.143 moles would need to be dissolved in 1 liter of water.