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.

257 grams h2o i think