question_answer

KBr is 80% ionized in solution. The freezing point of 0.4 molal solution of KBr is: \[{{K}_{f}}({{H}_{2}}O)=1.86K\text{ }kg/mole\]

A) 274.339K          

B) \[-1.339K\]

C) 257.3 K     

D) \[-1.339{}^\circ C\]

Correct Answer: D

Solution :

[d] \[\Delta {{T}_{f}}=i{{K}_{f}}m\] \[i=1+\alpha \] Here \[\alpha =0.8\] (80% ionization) =1.8 \[\Delta {{T}_{f}}=\left( 1.8 \right)\left( 1.86 \right)\left( 0.4 \right)=1.339\] \[{{T}_{f}}=-1.339{}^\circ C\]