Question

You collect a sample of hydrogen gas in an inverted buret by displacement of water at 23 degrees C. The buret could not be submerged deep enough in the water level in the water bath. The volume of gas in the buret was determined to be 32.60 mL.

a. If the current atmospheric pressure was 29.45" of Hg, what is the pressure of dry hydrogen in the buret? Show your work.

b. How many moles of hydrogen are in this sample? Show your work.

Answer 1

(a) From a knowledge or vapour pressure of water at different temperature (such a chart is easily available), we find the vapour pressure of water at 23°C to be 21.1 mm Hg.

The total atmospheric pressure is 29.45 mm Hg.

As per Dalton’s law,

P_total = P₁ + P₂ where P₁ and P₂ are the pressure of dry hydrogen and water vapour.

Therefore, pressure of dry hydrogen gas is (29.45 – 21.1) mm Hg = 8.35 mm Hg (ans).

(b) Now, we need to calculate the volume occupied by the dry hydrogen gas at 23°C. Assuming ideal behavior, we apply Boyle’s law as

P₁V₁ = P₂V₂ where P₁ = 29.45 mm Hg; V₁ = 32.60 mL; P₂ = 8.35 mm Hg. Therefore,

V₂ = (29.45 mm Hg)(32.60 mL)/(8.35 mm Hg) = 114.978 mL = 0.115 L

Now, we must convert the pressure of the gas in atmosphere and the temperature in Kelvin to use the ideal gas law.

We know that 760 mm Hg = 1 atm pressure.

Therefore, 8.35 mm Hg = 0.0110 atm.

Also, T = (273 + 23)K = 296 K.

Hence, we have

PV = nRT where n is the number of moles of dry hydrogen and R = 0.082 L-atm/mol.K is the universal gas constant.

Therefore, n = PV/RT = (0.0110 atm)(0.115 L)/(0.082 L-atm/mol.K)(276 K) = 5.59*10^-5 mol (ans)