Previous Year Questions [Chemistry] SOLUTIONS

 Previous Year Questions 

Chapter :Solutions 


1. What are collocation properties ?write Colligative property which is used to find the molecule mass of macromolecules .(2)

2. A solution of Glucose in the molar mass =  180 gram per mole in water has a boiling point of 100.20 degree Celsius. Calculate the freezing point of the same solution. Molal constants for water Kf and Kb 1.86 K kg per mole and 0.512 K kg per mole respectively (3) 

3.     Define the following terms:

1)     abnormal molar mass 

2)      van’t Hoff Factor (i)

4.     A 10% solution of sucrose in water has freezing point of 269.15 K. calculate the freezing point of 10% glucose in water if freezing point of pure water is 273.15 K.

5.  Calculate the freezing point of a solution containing 60 gram of glucose in 250 gram of water.[Kf =1.86 Kkg/mol ] 

6.     Give Reasons :

a)Measurement of osmotic pressure method is preferred for determination of molar masses of macromolecules such as proteins and polymers.

b)  Aquatic animals are more comfortable in cold water than in warm water.

7.Elevation of boiling point of 1M KCl solution is nearly double then that of 1M sugar solution. Calculate the freezing point of an aqueous solution containing 10.5 gram of magnesium bromide in 200 gram of water assuming complete dissociation of magnesium bromide

[ molar mass of magnesium bromide is equal to 184 g per mol and Kf for water is 1.8 6 K kg per mol.]

8. A 4% (w/w) solution of sucrose in water has a freezing point of 271.15 K. Calculate the freezing point of 5% glucose in water given freezing point of your water is equal to 273.15 K

9.Write the write 2 differences between an ideal solution and a non ideal solution.

10.Assertion : 0.1 M Solution of KCl has greater osmotic pressure than 0.1 m solution of glucose at same temperature

Reason : In solution KCl dissociates to produce more number of particles

AANS: osmotic pressure is a colligative property that depends upon the number of solute particles. Potassium chloride is a strong electrolyte which completely dissociates in the solution and number of particle increases while in case of glucose which is a weak electrolyte and does not dissociate in the solution so there is no change in the number of particles in the solution. Therefore kcl has more osmotic pressure then glucose. 

.[Option A

11.  State  RAOULT’s law for solution containing volatile components.what is the similarity between Raoult’s law and Henry's law

12.  Assertion : Osmotic pressure is a colligative property.

Reason : OSMOTIC PRESSURE IS DIRECTLY PROPORTIONAL TO MOLARITY

Ans: B 

13.a) Resolution contains 5.85 grams NCL per litre of solution it has an osmotic prasher of 4.75 atmosphere at 20:07 degrees Celsius. Calculate the degree of dissociation of NaCl in this solution.

b) State Henry's law. Why is air diluted with helium in the tanks used by scuba divers.

14.a) When 9.5 gram of F-CH2-COOH (molar mass = 78 g/mol) is dissolved in 500 g of water,the depression in freezing point is observed to be 1 degree Celsius.Calculate the degree of dissociation of ……………………….

Kf of water = 1.86 K kg/mol 

b)Give reasons : 

·       0.1 M KCl has higher boiling point than 0.1 M glucose

·       Meat is preserved for a longer time by salting

15. a) Why is boiling point of one MNCL solution more than that of 1M glucose solution

b) A non volatile solute X molar mass 50 gram per mole when dissolved in 78 gram of benzene reduced its vapour pressure to 90% calculate the mass of X dissolved in the solution

C) Calculate the boiling point elevation for a solution prepared by adding 10 g of magnesium chloride 200g of water assuming magnesium chloride is completely dissociated.[ Kb for water = 0.512 K kg/mol] 

16.a) Why van’t Hoff factor’s value for acetic acid in benzene close to 0.5 ?

b) Determine the osmotic pressure of a solution prepared by  dissolving 2.32 x10-2  g of potassium sulfate in 2 litres of solution at 25 degree Celsius assuming that potassium sulphate is completely dissociated.

[R= 0.082 L atm/K/mol,Molar mass of Potassium Sulphate = 174 g/mol]

c)     When 25.6 gram of sulphur was dissolved in 1000 grams of benzene the freezing point lowered by 0.512 Kelvin. Calculate the formula of sulphur for [benzene is equal to 5.12 Kelvin kg per mole atomic mass of sulphur is equal to 32 gram per mol]

17. Prove that Henry’s law is a special case of RAOULT’s law 

18. 30 gram of urea is dissolved in 846 gram of water. Calculate the vapour pressure of water for this solution if vapour pressure of water pure water add 298 Kelvin is 23.8 mm Hg.

19. Name the colligative property used for the determination of molar mass of polymers and proteins.

20. What is Henry's law give one application of it.

 

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