Question

A student used 0.0938 g Mg and collected 96.21 mL of H2 gas (their eudiometer tube is larger than the one we used) over water at 29.3 C on a day when the atmospheric pressure was 786.8 mmHg. The level of the solution in the tube was 12.6 mm above the level in the weaker when the reaction was complete. What was the experimental value for R in units of L mmHg mol-1 K-? How does this compare( calculate percent error) to the accepted value for R? .......I need step by step instuction please.

Answer 1

Vapor pressure of water at 29.3^oC = 30.3099 mm Hg
P_{height diff} = mm of solution * (density of solution / density of Hg)
P_{height diff} = 12.6 mm * (1.01 gm/mL / 13.534 gm/mL ) = 0.94029 mm Hg
P_H2 = atm pressure - P_{height diff} - VP_H2O
P_H2 = (786.8) - 0.94029 - 30.3099 = 755.5498 mmHg
V = 96.21mL = 96.21/1000 = 0.09621L
T = 29.3 + 273.15 = 302.45 K
Mg + 2HCl -----> MgCl₂ + H₂
From the above stoichiometry it is clear that 1 mol of Mg produces 1 mol of H₂
mol of Mg = 0.0938 gm / 24.3056 gm/mol = 3.8591*10^-3 mol
Therefore mol of H₂ = 3.8591*10^-3 mol = n
Use PV = nRT equation to find R
755.5498 mmHg * 0.09621 L = 3.8591*10^-3 mol * R * 302.45 K
R_exp = 62.27929 L mmHg mol^-1K^-1
The literature value is 62.36367 L mmHg mol^-1K^-1 = R_lit
Percent error = ( |R_lit - R_exp| / R_lit ) *100
% error = ( |62.36367-62.27929| / 62.36367) *100 = 0.13529 %