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Which of the following substances having concentration of aqueous solution 1% w/w, possesses higher boiling point ? $ (Molecular weight of Kcl, BaCl_2, glucose and Al_2(SO_4)_3 $ are 74.5, 208, 342 gm’mole respectively) (Assume that inonic solids dissociates completely in their aqueous solution)

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Explanation

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

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Molecular weight of biomolecules such as protein can be determined by method.

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Explanation

$ { 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 )

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At 353 K temperature, the Vapour pressure of pure liquids A and B are 600mm and 800 mm respectively. If mixture of liquids A and B boils at 353 K and 1 bar pressure, then mole proportion of B in percent is -

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90 gm glucose and 120 gm urea dissolved in 1.46 kg aqueous solution, then what will be the boiling point of the solution at 1 bar pressure?$(Kb =0.512 ^\circ C -kg –mole^{-1},$ molecular weight of glucose and urea are 180 and 60 gm/mole respectively)

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Explanation

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 $

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pH of 0.2M dibasic acid $ H_2A $ is 1.699; then, what will be its osmotic pressure at T K temperature ?

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Explanation

$ 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 $

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Boiling point of an aqueous soultion of $ 0.4m AlCl_3 is 100.7 ^\circ C; then what would be the pressure of ionization of AlCl_3 ? Kb – 0.512 ^\circ C - kg - mole-1.$

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The vapour pressure of homogenous mixture of 10 mole of liquid X and 30 mole of liquid Y at constant temperature is 550 mm. In this solution, 10 mole of liquid Y increases, hence, increase in vapour pressure is 10 mm. Then, find the vapor pressure of pure liquid X and Y at that temperature.

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Explanation

$ n_x = 10 , n _Y = 30 $ Total mok = 40 $ \therefore X_x = 0.25 X_y = 0.75 $ $ total vapour pressure = P = p_x + p-x $ $ \therefore n_X . P ^ 0_x + n_Y. p ^0 _Y = P $ $ \therefore 0.25 p^0 _X + 0.75p ^ 0 _Y = 550 ……(1) $ If mole of liquid y is in increases by 10 then its vapour pressure is increases by 10 mm $ \therefore n_X =10 , n_Y = 40 $ $ \therefore Total mole = 50 $ $ X_x = 0.2 , X_Y = 0.8 and total vapor pressure P = p_x + p_x = 560 mm $ $ \therefore 0.2 p ^0 _x = 0.8 p ^ 0 _Y =560 ......(2) $ By solving (1) and (2) we get $ p ^0 _x = 400 mm and p ^ 0 _Y = 600 mm $

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What amount of urea dissolved in 1 kg water at constant temperature, so that vapour pressure of the solution reduced by 2% ? ( M.W of urea = 60 gm/mole)

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Explanation

$ \triangle p = { 2p^0 \over 100 } \therefore { \triangle p \over p^ 0 } = { 1 \over 50} = X ( mole fraction of urea ) $ = { m \over 55.55 + m } $ $ \therefre m = 1.134 ( molality of urea ) $ $ \therefore mass of urea (W_2) = 1.134 \times 60 = 68 gm $

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What would be tne volume of 15% w/v and 5% w/v NaOH solution required to prepare 1 litre aqueous solution of 2M NaOH ? (M.w of Naoh = 40 gram/mole)

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Explanation

$ suppose V_1 liter 15 % W/V NaOH and V_1 liter 5% W/V NaOH solution is required to pre paec one litev 2 M NaOH solution$ $ 80 gm NaOh is required to prepare one liter 2m NaOH solution $ $ \therefore V_1 + V_2 = 1 liter ……..(1) $ $ 150 V_1 + 50 V_ 2 = 80 gm .......(2) $ By solving (1) and (2) $ we get V_1 = 300 ml and V_2 = 700 ml $

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At constant temperature, vapour pressure of an aqueous solution of 1.5 kg glucose decreases to 0.98% in comparision with vapour pressure of pure water then, what quantity of glucose in gram dissolved in the solution ? (Molecular weight of glucose = 180 gm/mole)

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