Solutions Question 372

Question: Benzene freezes at $ 5.50{}^\circ C $ . If the freezing point of 2.48 g of phosphorous in 100 g benzene is $ 4.48{}^\circ C $ , the atomicity of phosphorus in benzene is ( $ K_{f} $ (benzene) $ =5.12,K,kgmo{l^{-1}} $ ):

Options:

A) 1

B) 3

C) 4

D) 8

Show Answer

Answer:

Correct Answer: C

Solution:

Molecular mass of phosphorous $ =\frac{1000\times K_{f}\times W_{solute}}{\Delta T_{f}\times W_{benzene}} $

$ =\frac{1000\times 5.12\times 2.48}{1.02\times 100}=124.5={{(P)}_{x}} $ Or , $ x=\frac{124.5}{31}=4 $



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