NEET Practice Questions (MCQs) with Answers & Solutions

$ X = { m \over 55.55 + m } $ X = mole fraction of solute m = molality = 4.5 m $ \therefore X = { 4.4 \over 55.55 + 4.5 } = 0.075 $

$ Density of 98 \% w/w H_2 SO_4 = 1.8 gm/ ml $ w = 98 gm $ \therefore d = { w + w_o \over v } \therefore V = { w+W_o \over d } $ $ \therefore V = { 100 \over 1.8 } ml molarity ( m) = {1000 \times w \over M\ times V } $ $ = { 1000 \times 98 \times 9.8 \over 98 \times 100 } $ = 18 M

$ molality = { 100 \times molality \over ( 1000 \times density ) - (mol.mass of solute \times molarity ) }$

The sum of mole fractions of all components in a solution must be equal to 1. Given that the mole fractions of A, B, and C are 0.1, 0.2, and 0.4 respectively, the mole fraction of D can be calculated by subtracting the sum of these values from 1. (0.1 + 0.2 + 0.4 = 0.7), so the mole fraction of D is (1 - 0.7 = 0.3).

Mass of NaOH in 500 gm 25 % w/w NaOh solution $ = 5 \times 25 = 125 gm and mass of H_2 O = 5 \times 75 = 375 gm $ $ Mass of NaOh in 500 gm 15 % of w/w NaOH = 5 \times 15 = 75 gm and mass of H_2 O = 5 \times 85 = 425 gm $ $ Mass of NaOh in a mixed solution when both solutions are mixed W = 125 + 75 = 200 gm and mass of H_2 O = 375 + 425 = 800 gm $ $ Now molality of mixed solution = { 1000 \times W \over M \times W_0} = { 1000 \times 200 \over 40 \times 800} = 6.25 m $

molality of 25 % w/w NaOH $ = { 1000 \times w \over M \times W_o } = { 1000 \times 25 \over 40 \times 75 } = 8.33m $ $ molality of 15 % w/w of NaOh = { 1000 \times w \over M \times W_0} = { 1000 \times 15 \over 40 \times 85 } = 4.41 m $ $ When two different concentration solutions of same substances are mixed then conc of dil. solution \lt concentration of mixed solution \lt conc. of concentration soln $ $ \therefore 4.41 m \lt conc. ( molality ) of mixed solution \lt 8.33 in $