Question

A student reacted 0.0375 g of magnesium in a gas buret filled with water and HCl. The water temperature was 20.8C . The barometer in the lab read 758.1 mm Hg and 38.9 mL of gas were collected. Questions : What was the student's calculated value of R ? What is the percent deviation?

Answer 1

The reaction which is taking place in this set up is:

Mg + 2HCl ---> MgCl₂ + H₂(g)

Moles of Mg taken = Mass/MW = 0.0375/24.305 = 0.00154

So, moles of Hydrogen gas produced, n = Moles of Mg that reacted = 0.00154

Temp, T = 20.8⁰C = 293.8 K

Pressure, P = 758.1 mm Hg = 0.9975 atm

Volume, V = 0.0389 L

Putting values in the ideal gas equation we get:

PV = nRT

0.9975*0.0389 = 0.00154*R*293.8

Solving we get:

R = 0.08576 L.atm/(mol.K)

Accepted value = 0.0821 L.atm/(mol.K)

So,

% error = ((0.08576-0.0821)/0.0821) * 100 = 4.458%

Hope this helps !