Q: At 300 K , an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the height ( h ) of the solution (density =1.00 ~g ~cm ^ -3 ) where h is equal to 2.00 cm .

aligned & = g h=10^3 10 2 10^ -2 Pascal =200 Pascal \\ & =C R T \\ & 200= 2 M 1000 8.3 300 \\ & M=24900=2.49 10^4 ~g / mol \\ & X=2.49 aligned

Q: Vessel-1 contains w _2 ~g of a non-volatile solute X dissolved in w _1 ~g of water. Vessel- 2 contains w _2 ~g of another non-volatile solute Y dissolved in w _1 ~g of water. Both the vessels are at t

Vessel-I $ ( T _ b )_ I = i _ x w _2 M _ x 1 w _1 1000 K _ b $ $ M _ X =$ Molar mass of ' X ' Vessel-II $ ( T _ b )_ II = i _ Y w _2 M _ Y 1 w _1 1000 K _ b $ $M_Y=$ Molar mass of ' $Y$ ' $ aligned & ( T_b )_ I ( T _ b )_ II 100= i _ x i _ Y M _ Y M _ x 100 \\ & =1.2 100 80 100 \\ & =150 \% aligned $

Q: 50 mL of 0 . 2 molal urea solution (density = 1 . 012 g mL - 1 at 300 K ) is mixed with 250 mL of a solution containing 0 . 06 g of urea. Both the solutions w

Mole of urea = 0 . 2 Weight of urea = moles of urea × molar mass Weight of urea = 0 . 2 × 60 = 12 g Weight of solvent = 1000 g Weight of solution = 1012 g Volume of solution = Weight of solution density of solution ∴ Volume of solution = 1012 1 . 012 = 1000 ml ∵ 1000 ml solution contain 0 . 2 mole ∴ 50 ml solution contain = 0 . 2 × 50 1000 = 0 . 01 Mole of urea in other s

Q: Liquids A and B form ideal solution for all compositions of A and B at 25 ∘ C . Two such solutions with 0 . 25 and 0 . 50 mole fractions of A have the total vapor pressures of 0 . 3 and 0 . 4 ba

P Total = P A o X A + P B o X B ( i ) 0 .3 = ( P A o ) 1 4 + ( P B o ) 3 4 P A ∘ + P B ∘ = 1 . 2 ( ii ) 0 .4 = ( P A o ) 1 2 + ( P B o ) 1 2 P A ∘ + P B ∘ = 0 . 8 From equation (1) and equation (2) P B o = 0 .2 bar and P A o = 0 .6 bar

Question 1

In a solvent S , a compound B is partially dissociated into C and D as given below: B 2 C + 2 D B , C and D are non-volatile in nature. The molar mass of B is 10 times the molar mass of S . The standa

Accepted Answer

The elevation in boiling point is given by T_b = i K_b m. The ebullioscopic constant K_b is calculated as: K_b = R (T_b^ )^2 M_S 1000 H_ vap Substituting the given values (T_b^ = 400 K, H_ vap = 10R J mol^ -1 ): K_b = R (400)^2 M_S 1000 (10R) = 16 M_S For a dilute solution with 0.25\% (mass/mass) concentration, the mass of solute w_B = 0.25 g and the mass of solvent w_S 100 g. The molality m is: m = w_B 1000 M_B w_S = 0.25 1000 10 M_S 100 = 0.25 M_S Given T_b = 408 - 400 = 8 K, we substitute int

Question 2

Passage: Two volatile liquids A and B form an ideal solution. Consider a 5 molal solution of B in A inside a closed container having a total vapour pressure of 100 mm Hg at 300 K. The vapour pressure

Accepted Answer

For a 5 molal solution of B in A, there are 5 moles of B in 1000 g of A. Moles of A = 1000 50 = 20 mol. Mole fraction of B, x_B = 5 20 + 5 = 0.2 Mole fraction of A, x_A = 1 - 0.2 = 0.8 Using Raoult's law for the total vapour pressure: P_T = P_A^ x_A + P_B^ x_B 100 = 105 0.8 + P_B^ 0.2 100 = 84 + 0.2 P_B^ 0.2 P_B^ = 16 P_B^ = 80 mm Hg The molar volume of pure B in the liquid phase (V_ m,l ) is: V_ m,l = Molar mass of B Density of liquid B = 57 0.5 = 114 mL/mol = 0.114 L/mol Assuming pure B behave

Question 3

Passage: Two volatile liquids A and B form an ideal solution. Consider a 5 molal solution of B in A inside a closed container having a total vapour pressure of 100 mm Hg at 300 K. The vapour pressure

Accepted Answer

Molality of B in A is 5 m, which means 5 moles of B are dissolved in 1 kg (1000 g) of solvent A. Number of moles of A, n_A = 1000 50 = 20 mol Mole fraction of A in the liquid phase, x_A = n_A n_A + n_B = 20 20 + 5 = 0.8 Mole fraction of B in the liquid phase, x_B = 1 - 0.8 = 0.2 According to Raoult's law, the partial vapour pressure of A is: P_A = P_A^0 x_A = 105 0.8 = 84 mm Hg Given total vapour pressure, P_T = 100 mm Hg Partial vapour pressure of B, P_B = P_T - P_A = 100 - 84 = 16 mm Hg Mole f

Question 4

At 300 K , an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the height ( h ) of the solution (density =1.00 ~g ~cm ^ -3 ) where h is equal to 2.00 cm .

Accepted Answer

aligned & = g h=10^3 10 2 10^ -2 Pascal =200 Pascal \ & =C R T \ & 200= 2 M 1000 8.3 300 \ & M=24900=2.49 10^4 ~g / mol \ & X=2.49 aligned

Question 5

Vessel-1 contains w _2 ~g of a non-volatile solute X dissolved in w _1 ~g of water. Vessel- 2 contains w _2 ~g of another non-volatile solute Y dissolved in w _1 ~g of water. Both the vessels are at t

Accepted Answer

Vessel-I $ ( T _ b )_ I = i _ x w _2 M _ x 1 w _1 1000 K _ b $ $ M _ X =$ Molar mass of ' X ' Vessel-II $ ( T _ b )_ II = i _ Y w _2 M _ Y 1 w _1 1000 K _ b $ $M_Y=$ Molar mass of ' $Y$ ' $ aligned & ( T_b )_ I ( T _ b )_ II 100= i _ x i _ Y M _ Y M _ x 100 \ & =1.2 100 80 100 \ & =150 \% aligned $

Question 6

50 &#160; mL of 0 . 2 molal urea solution (density = 1 . 012 &#160; g &#160; mL - 1 at 300 &#160; K ) is mixed with 250 &#160; mL of a solution containing 0 . 06 &#160; g of urea. Both the solutions w

Accepted Answer

Mole of urea = 0 . 2 Weight of urea = moles of urea  &#215;  molar mass Weight of urea = 0 . 2 &#215; 60 = 12 &#160; g Weight of solvent = 1000 &#160; g Weight of solution = 1012 &#160; g Volume of solution  = Weight&#160;of&#160;solution density&#160;of&#160;solution &#8756; Volume of solution = 1012 1 . 012 = 1000 &#160; ml &#8757; &#160;&#160; 1000 &#160; ml solution contain 0 . 2 mole &#8756; 50 &#160; ml solution contain = 0 . 2 &#215; 50 1000 = 0 . 01 Mole of urea in other s

Question 7

An aqueous solution is prepared by dissolving 0 . 1  mole of an ionic salt in 1 . 8 &#160; kg of water at 35 &#176; C . The salt remains 90 % dissociated in the solution. The vapour pressure of t

Accepted Answer

Number of ions present per formula unit of ionic salt = x Van 't Hoff factor i = 1 + n - 1 &#945; &#160; &#160; = &#160; 1 + x - 1 &#215; 0 . 9 &#160; = &#160; 0 . 1 + 0 . 9 x (Assuming 90 % dissociation) &#8756; Relative lowering in vapour pressure = i &#215; Mole fraction of solute &#8658; 60 - 59 . 724 60 = i &#215; 0 . 1 1800 18 + 0 . 1 &#8658; &#160;&#160; 0 . 0046 = i &#215; 0 . 1 100 + 0 . 1 0 . 0046 &#8776; 0 . 9 x + 0 . 1 &#215; 0 . 1 100 &#8658; 0 . 9 x + 0 . 1 = 4 . 6 &#8658; &#16

Question 8

The boiling point of water in a 0.1 molal silver nitrate solution (solution A ) is x ^ C . To this solution A , an equal volume of 0.1 molal aqueous barium chloride solution is added to make a new sol

Accepted Answer

Given for solution  A m = 0 . 1 i=  1  for  AgNO 3 = 2 K b = 0 .5 &#160; Kg &#160; mol - 1 So  &#916;T b = T b &#8722; T b &#8728; = 1 &#160; K b m T b &#8722; 100 &#8728; C = 2 &#215; 0 .5 &#215; 0 .1 T b = 100 &#8728; C + 0 .1 = 100 .1 &#8728; C

Question 9

The boiling point of water in a 0.1 molal silver nitrate solution (solution A ) is x ^ C . To this solution A , an equal volume of 0.1 molal aqueous barium chloride solution is added to make a new sol

Accepted Answer

After adding  BaCl 2  solution in  AgNO 3 solution  AgCl  will be precipitate. Since volume of both solutions are equal let's take V = 1 &#160; L  of each. so initial moles of  AgNO 3 = 0 .1 &#160; mole initial moles of  BaCl 2 = 0 .1 &#160; mole 2 &#160; AgNO 3 + BaCl 2 &#8594; 2 AgCl &#8595; + Ba NO 3 2 2 AgNO 3 + BaCl 2 &#8594; 2 AgCl &#8595; + Ba NO 3 2 Intial &#160; moles 0 .1 0 .1 Change 0 .1 &#8722; 0 .05 + 0 .1 + 0 .05 Final &#160; moles 0 0 .0

Question 10

Liquids A and B form ideal solution for all compositions of A and B at 25 &#8728; C . Two such solutions with 0 . 25 and 0 . 50 mole fractions of A have the total vapor pressures of 0 . 3 and 0 . 4 ba

Accepted Answer

P Total&#160; = P A o X A + P B o X B ( i ) &#8196; 0 .3 = ( P A o ) 1 4 + ( P B o ) 3 4 P A &#8728; + P B &#8728; = 1 . 2 ( ii ) &#160; 0 .4 = ( P A o ) 1 2 + ( P B o ) 1 2 P A &#8728; + P B &#8728; = 0 . 8 From equation (1) and equation (2) P B o = 0 .2 &#8196; bar  and  P A o = 0 .6 &#8196; bar

Question 11

On dissolving 0.5 &#160; g of a non-volatile non-ionic solute to 39 &#160; g of benzene, its vapor pressure decreases from 650 &#160; m m &#160; H g to 640 &#160; m m &#160; H g . The depression of fr

Accepted Answer

&#916;P Ps = n solute n solvent 650 - 640 640 = 1 &#215; 0.5 &#215; 78 M &#215; 39 M solute = 64 &#8201;g &#916;T F = k f &#215; m &#8658; 5.12 &#215; 0.5 &#215; 1000 64 &#215; 39 &#916;T f = 1.0256

Solutions — Previous Year Questions & Solutions