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.
