A student collected hydrogen gas in a buret by displacing water at26.0 degrees C

Question

A student collected hydrogen gas in a buret by displacing water at26.0 degrees Celsius. The buret could not be submerged deepenough in a water bath to equalize the pressure. The waterlevel in the buret was 14.1 cm above the water level in the waterbath. The volume of the gas in the buret was determined to be33.6 mL.
a. If the atmospheric pressure was 753 torr, what is thepressure of the hydrogen in the buret? (The density of Hg is 13.6g/mL.)
b. How many moles of hydrogen did the studentcollect?
a. If the atmospheric pressure was 753 torr, what is thepressure of the hydrogen in the buret? (The density of Hg is 13.6g/mL.)
b. How many moles of hydrogen did the studentcollect?

Explanation / Answer

We Know that :      PTotal = Pdry + Paqueous tension       753 torr = Pdry + Paqu. tension         Paqu.tension = x g x h                               = 13.6 g / cc x 980 cm / s^2 x 14.1cm                                = 18792.48 N / m^2                                 = 140 .954 torr       Pdry   = 753 torr - 140.954torr                   =  612.045 torr         According to idealgas equation :           PV = nRT           612.045 / 760 atm x 33.6 / 1000 L   = nH2 x 0.0821 atm-L / mol-K x 299 K               number of moles of H2 = 0.00110moles         Paqu.tension = x g x h                               = 13.6 g / cc x 980 cm / s^2 x 14.1cm                                = 18792.48 N / m^2                                 = 140 .954 torr       Pdry   = 753 torr - 140.954torr                   =  612.045 torr         According to idealgas equation :           PV = nRT           612.045 / 760 atm x 33.6 / 1000 L   = nH2 x 0.0821 atm-L / mol-K x 299 K               number of moles of H2 = 0.00110moles

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