You collected a sample of hydrogen gas in an inverted buret by displacement of w

Question

You collected a sample of hydrogen gas in an inverted buret by displacement of water at 25 degree C. The buret could not be submerged deep enough in the water bath to equalize pressure. The water level in the buret was 13.80 cm above the water level in the water bath. The volume of gas in the buret was determined to be 38.75 ml. If the current atmospheric pressure was 30.10" of Hg, what is the pressure of dry hydrogen in the buret? Show work. How many moles of hydrogen are in this sample? Show work.

Explanation / Answer

Since the density of Hg is 13.6 times that of water,

1 torr = 1mm

Hg = 13.6mm water.

The water level in the buret is 13.80 cm above the water level.
13.8 cm water = 138mm water x [1torr / 13.6mm water] = 10.14 torr.

So pressure of H2 + pressure of water column = atmospheric pressure = 30.10 torr.
P H2 = 30.10 torr - 10.14torr = 19.96 torr.

A further correction needs to be made because the H2 gas also contains water vapor.
At 25o C the vapor pressure of water is 24.2 torr.
So pressure H2 + vapor pressure of water = 19.96 torr.
P H2 = 19.96 - 24.2 = - 4.24 torr.

Calculate the appropriate value for R.
PV = nRT
R = PV/nT
R = (30.10 torr)(22400 mL) / (1mole)(273K)

= 2469.74 torr mL / moleK

Finally, moles of dry H2:
PV = nRT
n = PV / RT

= (-4.24 torr)(38.75 mL) / [(2469.74 torr mL/moleK)*(299K)

= -164.3 / 738452.26

= -2.2249e-4 moles.

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