A) \[0.025MKN{{O}_{3}}<0.1MN{{H}_{2}}CSN{{H}_{2}}\] \[<0.05M\,BaC{{l}_{2}}<0.1MNaCl\]
B) \[0.1M\,NaCl<0.05M\,BaC{{l}_{2}}\] \[<0.1M\,N{{H}_{2}}CSN{{H}_{2}}<0.025M\,KN{{O}_{3}}\]
C) \[0.1M\,N{{H}_{2}}CSN{{H}_{2}}<0.1M\,NaCl\] \[<0.05M\,BaC{{l}_{2}}<0.025M\,KN{{O}_{3}}\]
D) \[0.025M\,KN{{O}_{3}}<0.05M\,BaC{{l}_{2}}\] \[<0.1M\,NaCl<0.1M\,N{{H}_{2}}CSN{{H}_{2}}\]
Correct Answer: B
Solution :
[b] Greater is the effective molarity \[\left( i\times C \right),\] higher the \[\Delta {{T}_{f}}\] value and lower the freezing point. (i) \[0.1\text{ }M\,NaCl=i\times C=2\times 0.1=0.2\] (ii) \[0.05M\,BaC{{l}_{2}}=i\times C=3\times 0.05=0.15\] (iii) \[0.1\text{ }M\,N{{H}_{2}}CSN{{H}_{2}}=i\times C=1\times 0.1=0.1\] (iv) \[0.025M\,KN{{O}_{3}}=i\times C=2\times 0.025=0.50\] Thus, order is (i) < (ii) < (iii) < (iv)
You need to login to perform this action.
You will be redirected in
3 sec
