Question

4. You collect a sample of hydrogen gas in an inverted buret by displacement of water at 230 °C. The buret could not be submerged deep enough in the water bath to equilize pressure. The water level in the buret was 14.65 cm above 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 (torr) of dry hydrogen in the buret? NG 2 b. How many moles of hydrogen are in this sample? NG 2
please help me with both A&B.

Answer 1

alut Piho Vapour Pressure of water at 23C = 21.1 mmHg PHbO = 21.1 month x 760 mong 1 atm Porto = 2.7763x16atm Pressure due to water column = egh = 1000 kg/m * 9.8 mm x 14.6535mx 2mm = 1435.7 Pax tone par = 1:4 17 x1ő? atm Atmospheric pressure = 29.45°Hg = 29.45X45.4 mmig. = 748.03 mm Hg X 760 mmHg = 0.9843 atm Atmospheric pressure = Pressure due to to. - Water column to the 0.9843 atm = 1.417x17?atm+ 2.7763x16matut Por PIK = 0.9843 -0.04193 Pitb = 0.9424 atm Round off | Pith = 0.94 atm RT Here, R=0.9424 atm, V=32.60 m) = 0.0326L R=0.082057 atw ma'E', T=2$C = 300k 0.9424 X 0.0326 0.082057 x 300 n=0.001248 mol [n= 1.2488103 mol ) or or m= 1.248 mmol