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Q.
The order of increasing freezing point of $C_2H_5OH, Ba_3(PO_4)_2, Na_2SO_4, KCl$ and $Li_3PO_4$ is
Solutions
Solution:
$ΔT_{f} ∝ i$
i for $C_{2}H_{5}OH = 1$, i for $Ba_{3}\left(PO_{4}\right)_{2} = 5$, i for $Na _{2}SO_{4} = 3,$ i for $KCl = 2$, i for $Li_{3}PO_{4} = 4$
Thus, depression in freezing point will be in order :
$C_{2}H_{5}OH < KCl < Na_{2}SO_{4} < Li_{3}PO_{4} < Ba_{3}\left(PO_{4}\right)_{2}$
and freezing point will be in order :
$Ba_{3}\left(PO_{4}\right)_{2} < Li_{3}PO_{4} < Na_{2}SO_{4} < KCl < C_{2}H_{5}OH$