A gas $A$ of molar mass $120 g/mol$ dissociates as $A (g) B (g) + 2 C (g)$ to the extent of 55\%. The average molar mass of the equilibrium mixture in $g/mol$ , to one decimal place, is

Numerical answer type, answer in g/mol, tolerance 0.05.

Answer: 57.1 g/mol

Per mole of A: moles at equilibrium

= (1 - α) + α + 2 α = 1 + 2 α = 2.10

M_{a vg} = 120/2.2 = 57.143

g/mol, i.e. 57.1.