JEE Main On 08 April 2018 Question 29
Question: For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point? [JEE Main Online 08-04-2018]
Options:
A) $ [\text{ Co(}{H_2}O{{)}_4}C{l_2}]\text{ Cl}\text{.2}{H_2}O $
B) $ [\text{ Co(}{H_2}O{{)}_3}C{l_3}]\cdot 3{H_2}O $
C) $ [\text{ Co(}{H_2}O{{)}_6}]\text{ C}{l_3} $
D) $ [\text{ Co(}{H_2}O{{)}_5}\text{Cl }]\text{ C}{l_2}\text{.}{H_2}O $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \Delta T _{f}=ik _{f}m $ $ {T _{f _{solvent}}}-{T _{f _{solution}}}=ik _{f}m $ For highest freezing point Vant hoff factor (i) should be lowest. for ionic compounds vant Hoff factors is equal to the number of discrete ions in a formula unit of the substance.
In First option 1 number of ions =2
In second option 2 Number of ions =1
In third option 3 Number if ions =4
In forth option 4 Number if ions =3 So, value if
i will be lowest in option 2 $ [\text{ Co(}{H_2}O{{)}_3}C{l_3}]\cdot 3{H_2}O $ So, option 2 will have lowest freezing point depression (ΔTf), so it will have highest freezing point.
