NEET Practice Questions (MCQs) with Answers & Solutions

$$ I = { \triangle T_{ob} \over \triangle T_{cal } } = { \pi_{ob} \over \pi_{cal} } = { M \over M_{ob} } $$ where M = Actual molecular mass of solute Mob = experimental molecular mass of solute For AB type ionoc compound $ I \leq 2 $ $ \therefore { M \over M_{ob} } \leq 2 $ $ 5 :2 \gt 2 :1 $

The substance forms trimmer in a solution due to association $$ \therefore { 1 \over 3 } \leq i\lt 1 $$ $$ \therefore { mkf \over 3 } \leq imKf \lt mKf $$ $$ \therefore { mkf \over 3 } \leq \triangle Tf \lt mKf $$ $$ \therefore -mKf \lt - \triangle Tf \leq - { mKf \over 3 } $$ $$ \therefore T ^ \circ _f - mKf \lt T ^ \circ _f - \triangle Tf \leq T ^\circ _f - { mKf \over 3 } $$ $$ \therefore T ^\circ _f - mKf \lt Tf \leq T ^ \circ _f - { mKf \over 3 } $$

If x molar solution of any substance is more concentrated than x molul solution so its molarity value is less than the molality value $ \therefore { molarity \over molarity } \lt 1 $ $$ Now { molarity \over molarity } = { 1000 \times w \over M \times V } \times { M \times W_0 \over 1000 \times W} $$
$$ \therefore { Molarity \over Molarity } = { w_0 gm \over V ml } \therefore { w_0 \over V} \lt {gm \over ml} $$ $$ Now { W_o \over V } = { W_0 w - w \over V} = { w_0 + W \over V} - { w \over v } = density of soln ( d { gm \over ml } ) - { w \over v } $$ $$ \therefore { W_0 gm \over V ml } = d { gm \over ml } + { molarity \times M \over 1000} $$ ( M = mol.mass of solute ) $$ \therefore d { gm \over ml} + { molarity \times M \over 1000} \lt 1 {gm \over ml } \therefore d { gm \over ml } \lt 1 { gm \over ml } + { molarity \times M \over 1000 } $$ $$ Here for 0.5 M H_2 SO_4 aqueous solution d \lt ( 1 + { 0.5 \times 98 \over 1000 } ) { gm \over ml } \therefore d \lt 1.049 { gm \over ml } $$

The boiling point of an aqueous solution of xm AlCl3 is given by Tb + i * kb, where i is the van't Hoff factor. For AlCl3, which is an ionic compound, i = 5 due to the dissociation of AlCl3 into Al3+ and 3Cl- ions. Hence, the correct option is Tb + 5 * kb.

$ { molarity \over molarity } = { 1000 \times W \over M \times V } \times { M \times W_0 \over 1000 \times W } $ $ \therefore { molarity \over molarity } = { W_0 \over V} } {gm \over ml} $ Now molarity = molality $ \therefore { W_0 \over V } = 1 { gm \over ml} $ $ \therefore d { gm \over ml} = 1 { gm \over ml} + { molarity \times M \over 1000} $ For urea molecular mass (M) = 60 gm/ml $ \therefore d = 1 + { molarity \times 60 \over 1000} $ $ \therefore d = { 50 + 3M \over 50 } $