Question

(Use the spaces provided for the answe 1. 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 How many grams of magnesium metal was (1) used? answer (2) If the vapor pressure of water had not been taken into account, what would have been the cal- culated value of R? answer (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 cal- culated value of R?

Answer 1

1) According to ideal gas law :

Pressure, P total = 735 torr

Pwater at 295 K = 19.8 torr

Pressure of dry hydrogen = Ptotal- Pwater = 735 - 19.8 torr = 715.2 torr = 0.941 atm ( 1 atm = 760 mmHg)

Volume, V = 35.6 ml

R = 82.1 ml atm/mol K

PV = nRT

0.941 atm x 35.6 ml = n x 82.1 ml atm/mol K x 295 K

33.50 mol = n x 24219.5

No. Of moles = 33.50 mol /24219.5 = 0.00138 moles

Mg(s) + 2HCl(aq) ----> MgCl₂(aq) + H_2(g)

Molar ratio of hydrogen and magnesium = 1:1

So moles of magnesium = 0.00138 moles

Molar mass of magnesium = 24.305 g/mol

Mass of magnesium = moles x molar mass = 24.305 g/mol x 0.00138 mol = 0.0336 gms

2) If vapor pressure of water is considered

Then pressure = 735 torr = 0.967 atm

PV = nRT

As pressure is increased

Let us consider n = 0.00138 moles

0.967 atm x 35.6 ml = 0.00138 moles x R x 295 K

R = 35.562/0.4071 ml atm/mol K = 87.35

R value increased if vapor pressure of water is not considered

3) if volume is 25.6 ml

As remaining parameters are constant

So V is directly proportional to R

V1/V2 = R1/R2

25.6 ml/35.6 ml = R1/82.1 ml atm/mol K

25.6 x 82.1/35.6 = R1

R1 = 2101.76/35.6 = 59.04 ml atm/mol K

Percent error = (calculated - exact value)/exact value ) x 100 = (82.1-59.04)/82.1) x 100 = 28.09 %