Question

A sample of 0.0883 g of M g, which has a molar mass of 24.31 g/mol, produces 82.1 mL of H 2 gas. The gas is collected over water at atmospheric pressure of 766.7 mm Hg at 22 oC, at which the vapor pressure of water is 19.8 mm Hg. What is the experimental value for the molar volume of the gas in L/mol?

Answer 1

Answer:

24 Lt/mol

Explanation:

Though we have many data here (such as molar mass of Mg, water vapor pressure, etc), we need to focus on data for H₂, which will help us to obtain the molar volume of this gas

Statement refers the following data for H₂:

V = 82.1 ml = 0.082 Lt

T = 22°C = 295 K

P atm = 766.7 mm Hg = 1.0089 atm (which is, the pressure for H₂ before being collected in water)

If we consider H₂ behaves as an ideal gas:

PV = nRT

Then we can move some terms from this ecuation :

V/n = RT/P so we can obtain the relation between V (volume) and n (N° of moles) for H₂, which is the experimental valur for the molar volume

Considering R = 0.082 Lt*atm/K*mol:

V/n = [(0.082 Lt*atm/K*mol)x295 K]/1.0089 atm

V/n = 23.97 Lt/mol (molar volume at this experiment conditions)