Determine the molecular mass of a gas where 3.87 g occupies 0.896 l at standard conditions.

Let's assume that the gas has ideal gas behavior.

Ideal gas law,

PV = nRT (1)

Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant (8.314 J mol ⁻¹ K⁻ ¹) and T is temperature in Kelvin.

n = m/M (2)

Where, n is number of moles, m is mass and M is molar mass.

From (1) and (2),

PV = (m/M) RT

By rearranging,

M = (mRT) / PV (3)

P = standard pressure = 1 atm = 101325 pa

V = 0.896 L = 0.896 x 10⁻³ m³

R = 8.314 J mol⁻¹ K⁻¹

T = Standard temperature = 273 K

m = 3.87 g = 3.87 x 10⁻³ kg

M = ?

By appying the formula,

M = (3.87 x 10⁻³ kg x 8.314 J mol⁻¹ K⁻¹ x 273 K) / 101325 pa x 0.896 x 10⁻³m³

M = 0.0967 kg

M = 96.7 g.

Hence, the molar mass of the gas is 96.7 g.