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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.$
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.
$ 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 $
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)
$ \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 $
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)
$ 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 $
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)
Boiling point of an aqueous solution of 0.5 m ionic solid substance is 100.5OC; then state the value of i ? $ (Kb = 0.512 ^\circ C -kg – mole^{-1}) $
Aqueous solution of substance boils at $ 100.5 ^\circ C at 1 bar pressure; then at what temperature it freezes ? (Kb = 0.512 ^\circ C -kg - mole^{-1}, Kf = 1.86 ^\circ C - kg – mole^{-1} ) $
1.4 m aqueous solution of a weak electrolyte AB2 ionizes 20%, then, state boiling point and freezing point of the solution respectively.
For a weak electrolyte solution, the boiling point elevation and freezing point depression depend on the degree of ionization. If the 1.4 m solution of AB2 ionizes 20%, then the effective molality is 1.4 x (1 + 0.2 x 2) = 2.24 m. Using the given values of Kb and Kf, the calculated boiling point and freezing point match option 2.
Solute substance in a 1.4 m aqueous solution associates by 25%, then, find the boiling point and freezing point of solution; where, solute exists as trimer in the solution; thus,n = 3. $ (Kb = 0.512 ^\circ C • kg • mole ^{-1} , Kf = 1.86 ^\circ C • kg • mole^{-1}) $
Molecular mass of a weak acid HA is 60gm/mo1 Ifs experimental molecular mass in its 0.7 M aqueous solution obtained from colligative properties is 50gm/mo1. Then calculate ionistion consteint of weak acid HA.
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