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 23.5 C. The buret could not be submerged deep enough in the water bath to equalize pressure. The water level in the buret was 11.45 cm above the water level in the water bath. The volume of gas in the buret was determined to be 36.6 mL.

a. if the current atmospheric pressure was 29.01'' of Hg, what is the pressure of dry hydrogen in the buret? show work.

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

Explanation / Answer

Part A

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

1 torr = 1mmHg = 13.6mmH2O.

11.45cm water = 114.5mm water x [1torr / 13.6mm water] = 8.42torr.

So pressure of H2 + pressure of water column = atmospheric pressure = 753 torr.

PH2 = 753torr - 8.42torr = 744.58torr.

A further correction needs to be made because the H2 gas also contains water vapor.

At 23.5 0C the vapor pressure of water is 21.717 torr.

So pressure H2 + vapor pressure of water = 742.6 torr.

PH2 = 742.6 - 21.717 = 720.88 torr.

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Part B

From ideal gas equation

PV = nRT

R = PV/nT

R = (760torr)(22400mL) / (1mole)(273K) = 62359 torr mL / moleK

Finally, moles of dry H2:

PV = nRT

n = PV / RT = (720.88torr)(36.6mL) / [(62359torr mL/moleK)*(299K)

n = 1.42*10-3 moles.

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