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Solutions MCQs for NEET — Chemistry Questions with Answers

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Difference of boiling point and freezing point of an aqueous solution of glucose is $ 104 ^ \circ C at 1 bar pressure; then what will be the molality of the solution ? (Kb = 0.513 ^\circ and K+” = 1.86 ^\circ C – kg - mole^{-1}) $

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Explanation

$ 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 $

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Difference of boiling point and freezing point of 0.2 m acetic acid prepared in benzene is 75.7 oC; then, state the value of Van’t Hoff factor i ? $ (For benzene, Kb = 2.65 OC – kg – mol^{–1}, Kf = 5.12 oC – kg – mol^{-1}, Tb = 80 oC, Tf = 5.5 ^\circ C) $

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Explanation

$ 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 $

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Difference in boiling point and freezing point of 10 kg aqueous solution of urea is 100.2372 OC; then what quantity of urea dissolved in the solution ? $ (Kb = 0.513 ^\circ and K+” = 1.86 ^\circ C – kg - mole^{-1} ) $

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Explanation

$ 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 $

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500 ml solution of HCl is prepared by dissolving 14.6 gm HCl in water. What will be the molarity of HCl in the solution ? (Molecular weight of HCl = 36.5 gm/mole )

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What would be the molality of the solution prepared by dissolving 60 gm NaOH in 1.5 kg water ? (Moleculea weight of NaOH = 40 gm/mole)

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What would be the molarity of $ 3.0 NH_2SO_4 $ solution ?

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What would be the molarity and normality of solution prepared by dissolving $ 19.6 gm H_2SO_4 in dissolved water to prepare 800 ml solution ? (Molecular weight of H_2SO_4 is 98 gm/mole) $

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What amount of $ H_2SO_4 required to prepare 2 litre of 0.5 NH_2SO_4 solution ? (Molecular weight of H_2SO_4 is 98 gm/mole) $

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Explanation

To prepare 2 litres of 0.5 M Hâ‚‚SOâ‚„ solution, we need 0.5 x 2 = 1 mole of Hâ‚‚SOâ‚„. Since the molecular weight of Hâ‚‚SOâ‚„ is 98 g/mol, the amount of Hâ‚‚SOâ‚„ required is 1 x 98 = 98 gm.

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What will be the concentration of solution prepared by dissolving 50 gm glucose in 200 gm water ?

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What is the normality of an aqueous solution of $ 0.5 Al_2(SO_4)_3 $ ?

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