Chemistry MCQs for NEET — Practice Questions with Answers

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 $

From graph $ p = p ^ \circ_A + ( p ^\circ_B - p ^ \circ _A ) X_B $ $ \therefore UR = QY + ( VW - QY ) QU $

$ { n \times \% W/ W \over molecular mass ( formula weight) } = X $
If value of x is highest than solution have highest boiling point ( n = no.of ions in a formula )

Mass of solvent in a solution wo = 1460 - ( 90 + 120 ) = 1250 gm Mole of glucose = 90 / 180 = 0.5 Mole of urea = 120 / 60 = 2 Total moles of solute in a solution = 0.5 + 2 = 2.5 $ molality = { 1000 \times n \over W_0} = { 1000 \times 2.5 \over 1250 } = 2.0 m $ $ \triangle Tb = mKb = 2 \times 0.512 = 1.024 ^ \circ C $ $ \therefore Tb = 100 + 1.024 = 101.024 ^ \circ C $

$ pH =1.669 \therefore [ H_3O ^+ ] = 0.02 M , [ H_2 A ] = 0.2 $ $ Degree of dissociation \alpha = { 0.02 \over 0.2 } = 0.1 = { i-1\over n -1 } ( n =3 ) $ $ \pi = iMRT = 1.2 \times 0.2 \times RT \therefore i=1.2 = 0.22 RT $