A) Nitrogen has an ATOMIC mass number of 14, but nitrogen gas consists of N₂ molecules, so the mass to use in this problem is 28 g/mol. Rates of effusion ∝ 1/√(mass), so
√(mass unknown) /√28 = (rate N₂ effusion)/(rate unknown effusion) = 1.59
∴ mass unknown = (1.59)²(28) = 70.78 g/mol
B) One possible gas that comes close for this mass is NF₃.
