Choose the pairs having identical value of van’t Hoff factor

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$0.05 M K_{4} [Fe {[CN]}_{6}]$ (50% degree of dissociation) and 0.05 M Mohr Salt ( 80% degree of dissociation)

0.05 M ${Na}_{3} {PO}_{4}$ ( 60% degree of dissociation) and 0.05 $K_{4} [Fe {(CN)}_{6}] (45 % degree of dissociation)$

0.2 M NaCl ( 80% degree of dissociation) and 0.2 M ${BaCl}_{2} (40 % degree of dissociation)$

0.01 M ${NaNO}_{3}$ ( 90% degree of dissociation ) and 0.01 M ${FeCl}_{3}$ (30% degree of dissociation)

answer is B, C, D.

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Detailed Solution

$\begin{array}{l} NaCl ⇌ {Na}^{+} + {Cl}^{-} \\ i = 1 + (2 - 1) (0.8) = 1.8 \\ {BaCl}_{2} ⇌ {Ba}^{2 +} + 2 {Cl}^{-} \\ i = 1 + (3 - 1) (0.4) = 1.8 \end{array}$
therefore (2) is the correct answer because both have identical values of i
$\begin{array}{l} {Na}_{3} {PO}_{4} ⇌ 3 {Na}^{+} + {PO}_{4}^{3 -} \\ i = 1 + (3) (0.6) = 2.8 \\ K_{4} [Fe [CN]_{6}] ⇌ 4 K^{+} + {[Fe (CN)_{6}]}^{4 -} \\ i = 1 + (4) (0.45) i = 1 + 1.8 = 2.8 \end{array}$