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Solutions Quiz-18

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IIT JEE exam which consists of JEE Main and JEE Advanced is one of the most important entrance exams for engineering aspirants. The exam is held for candidates who are aspiring to pursue a career in the field of engineering and technical studies. Chemistry is important because everything you do is chemistry! Even your body is made of chemicals. Chemical reactions occur when you breathe, eat, or just sit there reading. All matter is made of chemicals, so the importance of chemistry is that it's the study of everything..

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. A certain vessel X has water and nitrogen gas at a total pressure of 2 atm and 300 K. All the contents of vessel are transferred to another vessel Y having half the capacity of the vessel X. The pressure of N2 in this vessel was 3.8 atm at 300 K. The vessel Y is heated to 320 K and the total pressure observed was 4.32 atm. Assume that the volume occupied by the gasses in vessel is equal to the volume of the vessel. Calculate the following
Pressure of H2 O(g) in X at 320 K
  •  0.1
  •  0.2
  •  1.0
  •  2.0
Solution
Q2.A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the entropy of the system. A liquid vapourises to form a more disordered gas. When a solute is present, there is additional contribution to the entropy of the liquid due to increased randomness. As the entropy of solution is higher than that of pure liquid, there is weaker tendency to form the gas. Thus, a solute (non volaile) lowers the vapour pressure of a liquid, and hence a higher boiling point of the solution Similarly, the greater randomness of the solution opposes the tendency to freeze. In consequence, a lower temperature must be reached for achieving the equilibrium between the solid (frozen solvent) and the solution. The elevation in boiling point (∆Tb ) and depression in freezing point (∆Tf ) of a solution are the colligative properties which depend only on the concentration of particles of the solute and not their identity. For dilute solutions, ∆Tb and ∆Tf are proportional to the molarity of the solute in the solution.
Dissolution of a non-volatile solute into a liquid leads to
  •  A decrease of entropy
  •  An increase in tendency of the liquid to freeze
  •  An increases in tendency to pass into the vapour phase
  •  A decrease in tendency of the liquid to freeze
Q3.  Figure 2.36 (a) represents the distillation of mixture of liquid A and liquid B which gives both of pure liquid A and B. While Fig. 2.36(b) represents the azeotopic mixture of HNO3 and H2 O which distillation gives and azeotropic mixture and either of pure liquid. We can not separate both the pure liquid, i.e., H2 O and HNO3


What is the result of distilling a mixture of 50% HNO3 and 50% H2 O?
  •  Pure water and azeotropic mixture can be separated
  •  Pure H2 O and pure HNO3 can be separated
  •  Pure HNO3 and azeotropic mixture can be separated
  •  None of the above
Solution

Q4. Addition of non-volatile solute to a solvent always increases the colligative properties such as osmotic pressure, ∆P,∆Tb and ∆Tf. All these colligative properties are directly proportional to molality if solutions are dilute. The decrease in colligative properties on addition of non-volatile solute is due to increase in number of particles.
For different aqueous solutions of 0.1 N urea, 0.1 N NaCl, 0.1 N Na2 SO4 and 0.1 N Na3 PO4 solution at 27 ͦC the correct statements are : 1. The order of osmotic pressure is NaCl>Na2 SO4>Na3 PO4> urea 2. π=(∆Tb)/Kb ×ST for urea solution 3. Addition of salt on ice increases its melting point 4. Addition of salt on ice brings in melting of ice earlier
  •  2, 3, 4
  •  1, 2, 4
  •  1, 2, 3
  •  3, 4
Solution
Q5.Addition of a non-volatile solute to a solvent lowers its vapour pressure. Therefore, the vapour pressure of a solution (i.e.,V.P. of solvent in a solution) is lower than that of pure solvent, at the same temperature. A higher temperature is needed to raise the vapour pressure upto one atmosphere pressure, when boiling point is attained. However increase in b.p. is small. For example 0.1 molal aqueous sucrose solution boils at 100.05 ͦC. Sea water, an aqueous solution, which is rich in Na+ and Cl- ions, freezes about 1 ͦC lower than frozen water. At the freezing point of a pure solvent, the rates at which two molecules stick together to form the solid and leave it to return to liquid state are equal when solute is present. Fewer solvent molecules are in contact with surface of solid. However, the rate at which the solvent molecules leave the surface of solid remains unchanged. That is why temperature is lowered to restore the equilibrium. The freezing point depression in a dilute solution is proportional to molality of the solute.
An aqueous solution of 0.1 molal concentration of sucrose should have freezing point of (Kf=1.86):
  •  Hypotonic solution
  •  Hypertonic solution
  •  Isotonic solution
  •  Pure water
Solution

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SOLUTIONS Quiz-18
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