Question
Which of the following has been arranged in the increasing order of freezing point?
A.
$$0.025\,M\,KN{O_3} < 0.1\,M\,N{H_2}CSN{H_2}$$ $$ < 0.05M\,BaC{l_2} < 0.1M\,NaCl$$
B.
$$0.1M\,NaCl < 0.05M\,BaC{l_2}$$ $$ < 0.1M\,N{H_2}CSN{H_2} < 0.025M\,KN{O_3}$$
C.
$$0.1M\,N{H_2}CSN{H_2} < 0.1\,M\,NaCl$$ $$ < 0.05M\,BaC{l_2} < 0.025M\,KN{O_3}$$
D.
$$0.025M\,KN{O_3} < 0.05M\,BaC{l_2}$$ $$ < 0.1M\,NaCl < 0.1M\,N{H_2}CSN{H_2}$$
Answer :
$$0.1M\,NaCl < 0.05M\,BaC{l_2}$$ $$ < 0.1M\,N{H_2}CSN{H_2} < 0.025M\,KN{O_3}$$
Solution :
Greater is the effective molarity $$\left( {i \times C} \right),$$ higher the $$\Delta {T_f}$$ value and lower the freezing point.
$$\eqalign{
& \left( {\text{i}} \right)\,0.1\,M\,NaCl = i \times C = 2 \times 0.1 = 0.2 \cr
& \left( {{\text{ii}}} \right)\,0.05\,M\,BaC{l_2} = i \times C = 3 \times 0.05 = 0.15 \cr
& \left( {{\text{iii}}} \right)\,0.1\,M\,N{H_2}CSN{H_2} = i \times C = 1 \times 0.1 = 0.1 \cr
& \left( {{\text{iv}}} \right)\,0.025\,M\,KN{O_3} = i \times C = 2 \times 0.025 = 0.50 \cr
& {\text{Thus, order is (i)}}\,{\text{ < }}\,{\text{(ii)}}\,{\text{ < }}\,{\text{(iii)}}\,{\text{ < }}\,{\text{(iv)}} \cr} $$