Previous Year CBSE Questions From Chapter- 02 ( Solutions)
1)State Raoult’s law for solutions of volatile liquid components. Taking a suitable example, explain positive deviation from Raoult’s law.
2)A solution containing 8g of a substance in 100g of diethyl ether boils at 36.86oc, whereas pure ether boils at 35.60oc. Determine the molecular mass of the solute.( Foe ether, Kb= 2.02 K kg /mol.
3) State the law co-relating the pressure of a gas and its solubility in a liquid. State an application of this law.
4) Define the term ‘osmotic pressure’ . Describe how the molecular mass of a substance can be determined on the basis of osmotic pressure measurement.
5) Calculate the temperature at which a solution containing 54 g glucose in 250 g water will freeze. (Kf for water is 1.86 K Kg mol-1)
Year 2009
1)Define- (a) Mole Fraction (b) Van't Hoff Factor
2)100mg of a protein is dissolved in enough water to make 10.0ml of a solution. If this solution has an osmotic pressure of 13.3mm Hg at 25oc, what is the molar mass of the protein?( R=0.0821 L atm K-1mol-1 and 760mm Hg= 1 atm.
3) What is meant by - (a) Colligative properties (b) Molality of a solution
4. What conc of nitrogen should be present in a glass of water at room temp?
Assume a temp of 250C , a total pressure of 1 atmosphere and mol fraction of nitrogen in air of 0.78 ( KH for nitrogen = 8.42 × 10-7 M / mm Hg)
Year 2010
1)Define osmosis and osmotic pressure? What is the advantage of using osmotic pressure over other colligative properties
2)What mass of NaCl (58.5 g/mol) must be dissolved in 65g water to lower the freezing point by 7.5oC? Kf=1.86 K Kg/mol and I for Na Cl =1.86.
3) Differentiate between molarity and molality of a solution.
4) 15 g of an unknown molecular substance was dissolved in 450 g of water. The resulting solution freezes at -0.340C . What is the molar mass of the substance? ( Kf = 1.86 K Kg mol-1)
5) What mass of ethylene glycol ( molar mass = 62.0 g mol-1) must be added to 5.50 Kg of water to lower the freezing point of water from 00C to -10.00C ? ( Kf = 1.86 K Kg mol-1)
Year 2011
1)State- (i) )Raoult’s law in its general form in reference to solutions (ii)Henry’s law about partial pressure of a gas in a mixture.
2)A solution is prepared by dissolving 8.95mg of a gene fragment in 35ml of water has an osmotic pressure of 0.335 torr at 25oC. Calculate its molar mass if it is a non electrolyte.
3)What is reverse osmosis?
4)Difference between molarity and molality values for a solution. What is the effect of change in temperature on molarity and molality values.
5) What mass of NaCl (58.5 g/mol) must be dissolved in 65g water to lower the freezing point by 7.5oC? Kf=1.86 K Kg/mol and I for Na Cl =1.86
Year 2012
1) Define the following terms- (a) Mol Fraction (b) Ideal Soluion
2) 15 g of an unknown molecular substance was dissolved in 450 g of water. The resulting solution freezes at -0.340C . What is the molar mass of the substance? ( Kf = 1.86 K Kg mol-1)
3) Explain the following- (i) Henry’s Law about dissolution of gas in liquid (ii) Boilling point elevation constant for a solvent
4) A solution of glycerol (C3H8O3) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.420C . What mass of glycerol was dissolved to make this solution? ( Kb for water = 0.512 K Kg mol‑1)
Year 2013
(a) State Raoult’s law for a solution containing volatile components. How does Raoult’s law become a special case of Henry’s law?
(b) 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. Find the molar mass of the solute. (Kf for benzene = 5.12 K kg mol–1)
(c ) Define the following terms- (i) Ideal solution (ii) Azeotrope (iii) Osmotic pressure
(d) A solution of glucose (C6H12O6) in water is labelled as 10% by weight. What would be the molality of the solution? (Molar mass of glucose = 180 g mol–1)
Year 2014
a) Define the following terms- (i) Molarity (ii) Molal elevation constant (Kb)
b) A solution containing 15 g urea (molar mass= 60 g/mol) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass= 180 g/mol) in water. Calculate the mass of glucose present in one litre of its solution.
or
b) (i) What type of deviation is shown by mixture of ethanol and acetone? Give reason
b) (ii) A solution of glucose (molar mass = 180g/mol) in water is labelled as 10% (by mass). What would be the molality and molarity of the solution? (Density of solution + 1.2 g/mL)
Year 2015
(i) On mixing liquid X and liquid Y , the volume of the resulting solution increases. What type of deviation from Raoult’s law is shown by the resulting solution? What change in temperature would you observe after mixing liquids X and Y?
(ii) How can the direction of osmosis be reversed ? Write one use of reverse osmosis.?
(iii) A solution is prepared by dissolving 5 g of non volatile solute in 95 g of water. It has a vapour pressure of 23.375 mm Hg at 250C. Calculate the molar mass of the solute. ( Vapour pressure of pure water at 250C is 23.75 mm Hg)
Year 2016
(i) Write two differences between a solution showing positive deviation and a solution showing negative deviation from Raoult’s law
(ii) Calculate the freezing point of a solution when 3 g of CaCl2 ( M = 111 g mol-1) was dissolved in 100 g of water, assuming CaCl2 undergoes complete ionization. ( Kf for water = 1.86 K kg mol-1)
Year 2017
(i) A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water if the freezing point of pure water is 273.15 K. ( Molar mass of sucrose = 342 g mol-1, Glucose = 180 g mol-1)
(ii) Define - a) Molality ( m) b) Abnormal molar mass
(iii) 30 g of urea ( M = 60 g mol-1 ) is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg
( iv) Write two differences between ideal and non ideal solutions
Thank you for visiting the web page. I'll keep updating the page time to time.
If you find this helpful please share with your Chemy Friends.
MANISH KR SONI
0 Comments
Hi, Your feedback is highly appreciated.