Agaseous product has a mass of 2.34 g and occupies a volume of 0.854 l. the temperature in the laboratory is 302 k, and the air pressure is 1.04 atm. calculate the molar mass of the gas.

22.4 g/mol
44.0 g/mol
65.3 g/mol
86.9 g/mol

65.3 g/mol.

Explanation:

We can use the genral gas law of ideal gases:

PV = nRT,

where, P is the pressure of the gaseous compound in atm (P = 1.04 atm).

V is the volume of the gaseous compound in L (V = 0.854 L).

n is the no. of moles of the gaseous compound in mol.

R is the general gas constant (R = 0.082 L.atm/mol.K).

T is the temperature of the gaseous compound in K (T = 302 K).

∵ n = mass/molar mass.

∴ PV = (mass/molar mass)RT.

∴ Molar mass = (mass)RT/(PV) = (2.34 g)(0.082 L.atm/mol.K)(302 K)/(1.04 atm)(0.854 L) = 65.244 g/mol. ≅ 65.3 g/mol.

The right choice is (C) 65.3 g/mol

According to the ideal gas equation

PV = nRT

we have the n as the number of moles. Since we know that

moles = mass / molar mass

Mass is given to us as 2.34 g so arranging the above equation

PV = (mass/molar mass)RT

Putting the values

Molar mass = (mass)RT/(PV)

= (2.34 g)(0.082(302)/(1.04)(0.854 )

= 65.244 g/mol. OR approximately 65.3 g/mol.

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