The density of phosphorus vapour at 327 °C and 1 atm is $2.52 g {dm}^{- 3}$ . The molecular formula of phosphorus is

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P₂

P₃

P₄

answer is D.

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Detailed Solution

$M = \frac{ρ R T}{p} = \frac{(2.52 g {dm}^{- 3}) (8.314 k Pa {dm}^{- 3} K^{- 1} {mol}^{- 1}) (600 K)}{(101.325 k Pa)} = 124 g {mol}^{- 1}$ Number of P atoms $= \frac{124}{31} = 4$

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