Solutions Question 229
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 $
BETA
