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(B) The depression in freezing point is given by $\Delta T_f = i 	imes K_f 	imes m$. Since $K_f$ and $m$ are constant,$\Delta T_f \propto i$. Freezi

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$a-q, b-p, c-s, d-r$

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(B) The depression in freezing point is given by $\Delta T_f = i 	imes K_f 	imes m$. Since $K_f$ and $m$ are constant,$\Delta T_f \propto i$. Freezing point $T_f = T_f^0 - \Delta T_f = 273 - (i 	imes K_f 	imes 0.1)$. For sucrose,$i = 1$,so $\Delta T_f = 273 - 272 = 1 \ K$. Thus,$K_f 	imes 0.1 = 1$. Now calculate $i$ for each solution: $a$. $BaCl_2 ightarrow Ba^{2+} + 2Cl^-$,$i = 3$. $\Delta T_f = 3 	imes 1 = 3 \ K$. $T_f = 273 - 3 = 270 \ K$ $(q)$. $b$. $NaCl ightarrow Na^+ + Cl^-$,$i = 2$. $\Delta T_f = 2 	imes 1 = 2 \ K$. $T_f = 273 - 2 = 271 \ K$ $(p)$. $c$. $K_3[Fe(CN)_6] ightarrow 3K^+ + [Fe(CN)_6]^{3-}$,$i = 4$. $\Delta T_f = 4 	imes 1 = 4 \ K$. $T_f = 273 - 4 = 269 \ K$ $(s)$. $d$. $Al_2(SO_4)_3 ightarrow 2Al^{3+} + 3SO_4^{2-}$,$i = 5$. $\Delta T_f = 5 	imes 1 = 5 \ K$. $T_f = 273 - 5 = 268 \ K$ $(r)$. Matching: $a-q, b-p, c-s, d-r$.

Column-$I$ (Various solutions)	Column-$II$ (Their freezing point)
$a$. $0.1 \ M \ BaCl_2$ solution	$p$. $271 \ K$
$b$. $0.1 \ M \ NaCl$ solution	$q$. $270 \ K$
$c$. $0.1 \ M \ K_3[Fe(CN)_6]$ solution	$s$. $269 \ K$
$d$. $0.1 \ M \ Al_2(SO_4)_3$ solution	$r$. $268 \ K$

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