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A solution containing 28 g of phosphorus in 315 g ${CS}_{2}$ (b. p. 46.3° C) boils at 47.98° C. If $K_{b}$ for ${CS}_{2}$ is 2.34 $K kg {mol}^{- 1}$ '. The formula of phosphorus is (at. mass of P = 31).

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$P_{6}$

$P_{4}$

$P_{3}$

$P_{2}$

answer is B.

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Detailed Solution

$Δ T_{b} = 1.68 = 2.34 \times \frac{\frac{28}{M}}{315} \times 1000$

$\Rightarrow M = 124$

$31 \times x = 124$

$\Rightarrow x = 4$

Hence $P_{4}$

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