An evaluation of R was performed, following the procedure described in this modu

Question

An evaluation of R was performed, following the procedure described in this module. The barometric pressure was 736 torr, the temperature was 295 K, and the volume of hydrogen gas collected was 35.6 mL. The calculated value of R was 82.1 mL atm K^-1 mol^-1. (1) How many grams of magnesium metal were used? (2) If the vapor pressure of water had not been taken into account, what would have been the calculated value of R? (3) If the syringe volume had been incorrectly read, giving a calculated system volume of 25 6 mL, what would have been the percent error in the calculated value of R? (1) What is the volume of one mole of hydrogen gas at 273 K and 760 torr?

Explanation / Answer

The reaction is Mg+2HCl ---->MgCl2+ H2

Pressure =   736 Torr, Accounting for the vapor pressure of water at 295K (295-273=22 deg.c)

Vapor pressure of water at 22 deg.c =19.8 Torr

Partial pressure of hydrogen gas= 736-19.8= 716.2 Torr= 716.2/760 atm =0.942 atm

Volum, V= 35.6 ml, R= 82.1 mlatm/Kmole

Moles of H2 collected= PV/RT= 0.942* 35.6/(82.1* 295)= 0.0014

Hence from the reaction, moles of Mg used = moles of H2 generated= 0.0014

Atomic weight of Mg= 24, mass of Mg= moles* atomic mass= 0.0014*24=0.0336 gm

If vapor pressure of water is not accounted for, R= PV/(nT)= (736/760)* 35.6/(0.0014* 295)= 83.47 ml.atm/mole.K

When V is measured at 25.6 ml , R= (736/760)*25.6/(0.0014*295)= 60.02 ml.atm/mole.K

Percent error = 100*{(82.1-60.02)/ 82.1.}=26.89%

1 mole of any gas at standard conditions (273K and 760 Torr ) occupies

V= nRT/P=1*82.01*273/1 =22388 ml= 22.4 L

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