Which of the following aqueous solutions have been arranged in the order of their decreasing freezing points?

0.05 M K N O_{3} > 0.04 M C a C l_{2} > 0.140 M S u g a r > 0.075 M C u S O_{4}

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0.04 M C a C l_{2} > 0.05 M K N O_{3} > 0.140 M S u g a r > 0.075 M C u S O_{4}

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0.075 M C u S O_{4} > 0.04 M C a C l_{2} > 0.05 M K N O_{3} > 0.140 M S u g a r

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0.140 M S u g a r > 0.075 M C a S O_{4} > 0.04 M C a C l_{2} > 0.05 M K N O_{3}

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Solution

The correct option is A $0.05 M K N O_{3} > 0.04 M C a C l_{2} > 0.140 M S u g a r > 0.075 M C u S O_{4}$
We know that,
$Δ T_{f} = i K_{f} m$
So more the $Δ T_{f}$ value, lower the freezing point.
$Δ T_{f} \propto i \times m$ (Since $K_{f}$ is same for all the solutions given)
$i \times m$ for,
$0.05 M K N O_{3}$ = $0.05 \times 2 = 0.1$
$0.04 M C a C l_{2}$ = $0.04 \times 3 = 0.12$
$0.140 M S u g a r$ = $0.140 \times 1 = 0.140$
$0.075 M C u S O_{4}$ = $0.075 \times 2 = 0.15$

Since $i \times m$ is lowest for $0.05 M K N O_{3}$ , it has the highest freezing point, followed by $0.04 M C a C l_{2}$ , $0.140 M s u g a r$ and the lowest freezing point will be for $0.075 M C u S O_{4}$ .

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