CBSE Class 12 Chemistry 2014 Outside Delhi Set 1 Solved Paper

© examsnet.com
Question : 28
Total: 30
(a) Define the following terms:
(i) Molarity
(ii) Molal elevation constant (kb)
(b) A solution containing 15g urea (molar mass =60 gmol1 ) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass =190gmol1 ) in water. Calculate the mass of glucose present in one litre of its solution.
OR
(a) What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
(b) A solution of glucose (molar mass =180gmol1 ) in water is labelled as 10% (by mass). What would be the molality and molarity of the solution? (Density of solution =1.2gmL1 )
Solution:  
(a) (i) Molarity of a solution is defined as the number of moles of solute present in one litre of the solution.
(ii) Molal elevation constant (Kb) is defined as the elevation of the boiling point of a solution when one mole of a non-volatile solute is dissolved in one kilogram of a volatile solvent.
(b) Given; Mass of urea, WB=15g
Molar mass of urea, Murea =60g
The solution of urea in water is isotonic to that of glucose solution.
πurea =πglucose
Curea RT=Cglucose RT
nurea RT
V
=
nglucose
V
RT

15
60
=
Wglucose
180

Wglucose =
15×180
60

Wglucose =45g
OR
(a) According to Raoult's law, the partial pressure of a component is the product of vapour pressure of pure solvent and mole fraction of that component. When a solution shows deviation from Raoult's law over the complete range of concentration, the solution is known as a non-ideal solution.
The vapour pressure of the non-ideal solution can be higher or lower than the vapour pressure predicted by Raoult's law.
If the vapour pressure of the non-ideal solution is higher than the vapour pressure predicted by Raoult's law, the deviation is known as positive deviation.
If the vapour pressure of the non-ideal solution is lower than the vapour pressure predicted by Raoult's law, the deviation is known as negative deviation.
The reason for the deviation is molecular interactions, AB<AA,BB for positive deviation
Here, A is the solute and B is the solvent so, A-B shows the interaction between solute and solvent.
(b) The number of moles of glucose
=
10
180
=0.056mol

Molality of solution =
0.056mol
0.09kg
=0.62m

If the density of the solution is 1.2gmL1, then the volume of the 100g solution can be given as,
100g
1.2mL1
=83.3mL

=83.33×103L
Molarity of solution =
0.056mol
83.33×103L
=0.67M
© examsnet.com
Go to Question: