Chemistry MCQs for NEET — Practice Questions with Answers

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 } $

$ T_b - T_f = 104 $ $ T_b ^ o + \triangle T_b - ( T_f ^ \circ - \triangle T_f ) = 104 $ $ \therefore 100 + \triangle T_b - (0 - \triangle T_f ) = 104 $ $ \therefore \triangle T_b + \triangle T_f = 4 $ $ For glucose i=1 \therefore mKb + mKf = 4 \therefore m = { 4 \over (Kb+Kf) } = { 4 \over (0.513 + 1.86 ) } \therefore m = 1.68 $

$ T_b - T_f = 100.2372 $ $ T_b ^ o + \triangle T_b - ( T_f ^ \circ - \triangle T_f ) = 100.2372 $ $ \therefore 100 + \triangle T_b - (0 - \triangle T_f ) = 100.2372 $ $ \therefore \triangle T_b + \triangle T_f = 0.2372$ $ For urea I = 1 \therefore mKb + mKf = 0.2372 $ $ \therefore m = { 0.2372 \over Kb + Kf } = { 0.2372 \over 0.513 + 1.86 } $ $ \therefore m = 0.1 $

$ T_b - T_f = 75.7 $ $ T_b ^ o + \triangle T_b - ( T_f ^ \circ - \triangle T_f ) = 75.7 $ $ \therefore 80 + \triangle T_b - (5.5 - \triangle T_f ) = 75.7 $ $ \therefore \triangle T_b + \triangle T_f = 1 .2 $ $ \therefore imKb + imKf = 1.2 \therefore i = { 1.2 \over m (Kb+Kf) } = { 1.2 \over 0.2 (2.65 + 5.12 ) } \therefore i = 0.77 $