Which of the following has been arranged in the increasing order of freezing point?

Which of the following has been arranged in the increasing order of freezing point?
  1. $0.025 \mathrm{M~} \mathrm{KNO}_{3} < 0.1 \mathrm{M~} \mathrm{NH}_{2} \mathrm{CSNH}_{2} < 0.05 \mathrm{M~} \mathrm{BaCl}_{2} < 0.1 \mathrm{M~} \mathrm{NaCl}$
  2. $0.1 \mathrm{M~} \mathrm{NaCl} < 0.05 \mathrm{M~} \mathrm{BaCl}_{2} < 0.1 \mathrm{M~} \mathrm{NH}_{2} \mathrm{CSNH}_{2} < 0.025 \mathrm{M~} \mathrm{KNO}_{3}$
  3. $0.1 \mathrm{M~} \mathrm{NH}_{2} \mathrm{CSNH}_{2} < 0.1 \mathrm{M~} \mathrm{NaCl} < 0.05 \mathrm{M~} \mathrm{BaCl}_{2} < 0.025 \mathrm{M~} \mathrm{KNO}_{3}$
  4. $0.025 \mathrm{M~} \mathrm{KNO}_{3} < 0.05 \mathrm{M~} \mathrm{BaCl}_{2} < 0.1 \mathrm{M~} \mathrm{NaCl} < 0.1 \mathrm{M~} \mathrm{NH}_{2} \mathrm{CSNH}_{2}$

Solution

Greater is the effective molarity $(1 \times C)$, higher the $\Delta \mathrm{T}_{\mathrm{f}}$ value and lower the freezing point. (i) $0.1 \mathrm{M~} \mathrm{NaCl}=\mathrm{i} \times \mathrm{C}=2 \times 0.1=0.2$ (ii) $0.05 \mathrm{M~BaCl}_{2}=\mathrm{i} \times \mathrm{C}=3 \times 0.05=0.15$ (iii) $0.1 \mathrm{M~} \mathrm{NH}_{2} \mathrm{CSNH}_{2}=\mathrm{i} \times \mathrm{C}=1 \times 0.1=0.1$ (iv) $0.025 \mathrm{M~} \mathrm{KNO}_{3}=\mathrm{i} \times \mathrm{C}=2 \times 0.025=0.50$ Thus, order is $(\mathrm{i}) < (\mathrm{ii}) < (\mathrm{iii}) < (\mathrm{iv})$ ~

Asked in: JEE-TOPICTESTS-CHEMISTRY

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