what will the pressure be of this problem? will it be 30.25in Hg×25.4mm/in =768.

Question

what will the pressure be of this problem?
will it be 30.25in Hg×25.4mm/in =768.4mmhg×1 atm/760mmhg=1.010986 atm
OR
30.25inHg×25.4mm/in =768.4mmhg -23.8 mmhg (since the vapor pressure of water in eudiometer is 23.8mmhg when the T is 25.0C) =744.55mmhg×1atm/760mmhg=0.9797 atm

does this problem need me to minus 23.8mmhg? or does this problem not need to include the 23.8 mmhg in the calculations and the pressure is simply 1.010986 atm

4. In an experiment undertaken much like yours, a student reacted 0.0395g of magnesiu in a eudiometer filled with water and HCI. The water temperature was 25.0 C. The barometer in the lab read 30.25 in. Hg and 48.20mL of gas was collected. a. What was the student's calculated value of R? b. What is the percent error? c. What is one possible source of this error?

Explanation / Answer

Part a

Barometric pressure = 30.25 in Hg x 25.4mm/in

= 768.35 mmHg

Vapor pressure of water 25°C = 23.8 mmHg

Pressure of dry gas = Barometric pressure - Vapor pressure

= 768.35 - 23.8

= 744.55 mmHg x 1atm/760mmHg

= 0.9797 atm

Volume of gas V = 48.20 mL x 1L/1000 mL

= 0.04820 L

Mg + 2 HCl = MgCl2 + H2

Moles of H2 = moles of H2 = 0.0395g /24.305 g/mol

= 0.001625 mol

Temperature T = 25 + 273 = 298 K

Gas constant R = PV/nT

= 0.9797 atm x 0.04820 L/0.001625 mol x 298 K

= 0.09751 L-atm/mol-K

Part b

% error =( experimental value - actual value) *100/actual value

= (0.09751 - 0.0821)*100/0.0821

= 18.77 %

Part C

Possible source of error

Barometer could have detected the high pressure rather than the original

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