Question

34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °c and 0.1 bar pressure. What is the molar mass of phosphorus?

Answer 1

Given,

p = 0.1 bar

V = 34.05 mL = 34.05 × 10–3 L = 34.05 × 10–3 dm3

R = 0.083 bar dm3 K–1 mol–1

T = 546°C = (546 + 273) K = 819 K

From the gas equation PV = w. RT / M, we get

M = w. RT/ Pv    ……….(1)

Changing the given values in the equalization (1), we see

M = (0.0625 / 0.1 x 34.04) X 82.1 X 819 = 124.75 g/mol

Therefore, the molar quantity of phosphorus is 124.75 g mol–1.