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TEST PAPER ,CBSE NCERT SAMPLE Paper

Q.1 :- Calculate the molarity of 25% by weight of aqueous solution of NaOH.


Q.2 :- Calculate the molarity of 6g of CH3COOH solution of 600 m.l. volume.


Q.3 :- How will you express the mass percentage of a component of solution?


Q.4 :- What does it mean by 10% urea in water by mass?


Q.5 :- What does it mean by volume percentage of solution?


Q.6 :- What does it mean by 20% acetic aqueous solution?


Q.7 :- If you are given with 10% by weight glucose solution. How will you calculate


mole fraction of glucose in solution?


Q.8 :- What is solubility?


Q.9 :- On which factors does solubility depend?


Q.10 :- What is the effect of pressure on solubility of gases in liquids?


Q.11 :- What is the effect of temperature on solubility of gas in liquids?


Q.12 :- How the solubility of a gas in liquids is an exothermic process?


Q.13 :- Give an equation for relative lowering of vapour pressure?


Q.14 :- Why the vapour pressure of solvent in solution is less than that of the pure


solvent?


Q.15 :- Derive the formula for relative lowering of vapour pressure?


Q.16 :- Give the formula for elvation of boiling point?


Q.17 :- Give the formula for depression in freezing point?


Q.18 :- What is a semi-permeable memberane?


Q.19 :- What is osmosis?


Q.20 :- Why alcohol is soluble in water?


Q.21 :- What type of bonding is there in a solution of methanol and acetone?


Q.22 :- What is the chemical formula of ethylene glycol?


Q.23 :- Now tell whether it will be soluble in water or not?


Q.24 :- Write the molecular formula of Aspirin?


Q.25 :- Now tell whether Aspirin would soluble in acronitrile (CH3CN) or not?


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ANSWERS

Ans :-1. Weight of NaOH = 25g

Weight of solvent = 75g
25 5
Number of moles of NaOH = =
40 8
5
75 g of solvent contains = moles of NaOH
8
5
1g of solvent contains = moles of NaOH
8 X 75
5X1000 25
1000 g of solvent contains = = moles of NaOH
8X75 3
25
Hence, molality =
3 6 1
Ans :- 2. Number of moles of CH3COOH = =
60 10
1
600ml.. of solution contains = moles of CH3COOH
10
1
1 ml. of solution contains = moles of CH3COOH
10 X 600
1 X 1000 1
1000 ml of solution contains = = moles of CH3COOH
10 X 600 6


1
Hence molarity =
6
Mass of the component in solution X 100
Ans :- 3. Mass % of a component =
Total mass of solution
Ans :- 4. It means that 10g of urea is dissolved in 90g of water. So, that the overall weight

of solvent is 100g.
Volume of component X 100
Ans :- 5. Volume % =
Total volume of solution

Ans :- 6. 20 m.l. of acetic acid is dissolved in water such that the total volume of solution

is 100 m.l.


Ans -7 mole fraction = number of mole of glucose
Total no. of mole in solution

Now, we have 18g of glucose in 90g of water
Wt. of glucose
No. of moles of glucose =
Mol. wt. of glucose
18 1
= =
180 10
82 41

No. of moles of water = =
18 9

1/10 9
Mole fraction of glucose = =
1/10+41/9 419

Ans -8 Solubility of substance is its maximum amount that can be dissolved in a specified amount of solvent.

Ans-9 (i) Nature of solute and solvent
(ii) Temperature.
(iii) Pressure .

Ans -10 The solubility of a gas in liquid increases with increase in pressure.

Ans-11 Solubility of a gas in liquid decreases with rise in temperature because it
is an exothermic process.

Ans-12 Because this dissolution can be considered similar to condensation as gas gets
Converted into liquid which involves evolution of heat.

Ans-13 pl = pi – pl = x2
Pi pi

Ans -14 in pure solvent, the particles at the surface are all of solvent which are available for evaporation. So vapour pressure is more but in solution, solute is also there which is non- volatile so some molecules are of solute and some are of solvent. Since, number of solute and some are solvent Since, number of solvent particles at the surface is less and v.p. decreases.


Ans- 15 Let,-
P l = v.p. of solution
X l = mole fraction of solvent
Pi = v.p. of pure solvent

P l = pi x l
if pl = reduction of v.p.
pl = pi – pl
= pi – pi x l
= pi ( 1-xl )
= pi x 2
pl
pi = x2 Relative lowering of v.p. is equal to mole fraction of solute.

Ans- 16 Tb = Kb m

Tb = Increase in b.p. when a solute is added to solvent.
Kb = molal elevation constant it depends upon the nature of solvent.
M = molality.
Ans -17 Tf = kf m
Tf = decrease in freezing point
Kf = freezing point depression constant. It depends upon nature of solvent.
m = molality.

Ans -18 It is a memberance which allows to pass the particles only of a particular size selectively.All particles can not pass through it.
Ans 19 :- When a solute and a solvent are separated by a semi-permeable memberance, the solvent flows towards the solution.The process of flow of solvent is called osmosis.
Ans -20 The -OH group of alcohol forms hydrogen bonding with water molecules hence it is soluble.
Ans-21 Hydrogen bonding.
Ans-22 CH2 – OH
CH2 – OH
Ans- 23 Yes.
Ans-24 C9 H8 O4
Ans -25 Yes.S

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