SOLUTIONS Quiz-17

Solutions Quiz-17

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This section contain(s) 10 paragraph(s) and based upon each paragraph, multiple choice questions have to be answered. Each question has atleast 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct..

Q1. The vapour pressure of the solution is comparatively smaller than that of the pure solvent (Raoult’s law). Consequently, the temperature at which the vapour pressure of solution becomes equal to the external pressure, will be greater than that of pure solvent, rising its boiling point. Since, the decrease in vapour pressure is directly proportional to the amount fraction of the solute in the solution, it is therefore expected that the corresponding increase in the boiling points also depends on the amount fraction of the solute in the solution
When a non-volatile solute is dissolved in a solvent, the relative lowering of vapour pressure is equal to

•  Concentration of the solute in gram 100 mL
•  Mole fraction of solute
•  Mole fraction of solvent
•  Concentration of the solute in gram per litre

Solution

Q2.The variation of vapour pressure of the solvent and that of the solution with temperature are given by the respective solvent-vapour and solution-vapour curves of the phase diagram

For a given value of the external pressure, the pure solvent will boil at temperature T_b^* and at temperature T_b let p_ext be equal to p^*, the vapour pressure of pure solvent Applying the Clausius-Clapeyron equation to the solution vapour equilibrium for the two values of p,T_b^* and p^*,T_b we have In p^*/p=(∆_vap H₁m)/R (1/(T_b^* )-1/T_b ) =(∆_vap H₁m)/R (∆T_b)/(T_b^* T_b )
The phase diagram for the pure solvent and solution are recorded below. The quantity indicated by L in the figure is

•  ∆p
•  k_b∙m
•  k_f m
•  m

Q3.  An aqueous solution freezes at 272.4 K while pure water freezes at 273 K. Given, K_f=1.86 K kg mol^(-1),K_b=0.512 K kg mol^(-1) and vapour pressure of water at 298 K = 23.756 mm of Hg. Determine the following
Molality of the solution is

•  0.322
•  0.222
•  0.413
•  0.5

Solution

Q4. A solution of sucrose (molar mass =342) is prepared by dissolving 68.4 g in 1000 g of water. Calculate
The vapour pressure of solution at 293 K

•  0.0229
•  0.4
•  0.6
•  0.9

Solution

Q5.The osmotic pressure π depends on the molar concentration of the solution (π=CRT). If two solutions are of equal solute concentration and, hence, have the same osmotic pressure, they are said to be isotonic. If two solutions are of unequal osmotic pressures, the more concentrated solution is said to be hypertonic and the more diluted solution is described as hypertonic. Osmosis is the major mechanism for transporting water upward in the plants. Answer the following questions:
A plant cell shrinks when it is kept in:

•  Hypotonic solution
•  Hypertonic solution
•  Isotonic solution
•  Pure water

Solution

Q6. The electrolyte solutions show abnormal colligative properties. To account for this effect we define a quantity called the Van’t Hoff factor given by i=(Actual number of particles in solution after dissociation)/(Number of formula units initially dissolved in solution) i=1 (for non-electrolytes) i> 1 (for electrolytes, undergoing dissociation) i< 1 (for solutes, undergoing association) Answer the following questions:
Benzoic acid undergoes dimerization in benzene solution. The Van’t Hoff factor i is related to the degree of association α of the acid as

•  i=2-α
•  i=1+(α⁄3)
• i=1-(α⁄2)
•  i=1+(α⁄2)

Solution

Q7.

Compartments A and B have the following combinations of solution:

Indicate the number of solutions which is/are isotonic

•  1 only
•  3 only
•  4 only
•  2 only

Solution

Q8.The boiling point elevation and freezing point depression of solutions have a number of practical applications. Ethylene glycol (CH₂ OH-CH₂ OH) is used in automobile radiators as an antifreeze because it lowers the freezing point of the coolant. The same substance also helps to prevent the radiator coolant from boiling away by elevating the boiling point. Ethylene glycol has low vapour pressure. We can also use glycerol as an antifreeze. In order for the boiling point elevation to occur, the solute must be non-volatile, but no such restriction applies to freezing point depression. For example, methanol (CH₃ OH), a fairly volatile liquid that boils only at 65℃, is sometimes used as an antifreeze in automobile radiators
Which of the following is a better reagent for depression in freezing point but not for elevation in boiling point?

•  CH₃ OH
•  
  
•  
  
•  C₆ H₁₂ O₆

Solution

Q9.A solution M is prepared by mixing ethanol and water, the mole fraction of ethanol in the mixture is 0.9 Given : Freezing point depression constant of water (K_f^water )=1.86 K kg mol^(-1) Freezing point depression constant of ethanol (K_f^ethanol )=2.0 K kg mol^(-1) Boiling point elevation constant of water (K_b^water )=0.52 K kg mol^(-1) Boiling point elevation constant of ethanol (K_b^ethanol )=1.2 K kg mol^(-1) Standard freezing point of water = 273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethanol = 40 mm Hg Molecular weight of water = 18 g mol^(-1) Molecular weight of ethanol = 46 g mol^(-1) In answering the following questions consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative
The freezing point of the solution M is

•  268.7 K
•  268.5 K
•  150.9 K
•  268.7 K

Solution

Q10.Properties such as boiling point, freezing point, and vapour pressure of a pure solvent change when solute molecules are added to get homogenous solution. These are called colligative properties Answer the following questions:
I 0.001 m NaCl II 0.001 m urea III 0.001 m MgCl_2 IV 0.001 m CH_3 COOH Increasing order of boiling points

•  (ii) < (iv) < (i) < (iii)
•  (iv) < (i) < (ii) < (iii)
•  (iii) < (ii) < (i) < (iv)
•  (i) < (ii) < (iii) < (iv)

Solution

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