2. SOLUTION
VSA (1Mark)
1.Why is glycol and water mixture used in car radiators in cold countries?
Ans: Ethylene Glycol reduces the freezing point of water. Due to this, the coolant
in radiators will not freeze.
2. Give reason When 30 ml of ethyl alcohol and 30ml of water are mixed, the volume
of resulting solution is more than 60ml.
Ans.Solution of water and ethyl alcohol shows positive deviation and hence 𝛥𝑉 > 0
3. Cutting onions taken from fridge is more comfortable than cutting onions lying at room
Temperature. Explain.
Ans. At lower temperature, vapor pressure is low, hence less vapors of tear-producing
chemicals is formed.
4.Two liquids A &B on mixing produce a warm solution. Which type of deviation from
Raoult’s Law is expressed?
Ans. This exhibits that dissolution of A & B is an exothermic process. Hence, it shows
negative deviation from Raolt’s Law.
5.Two liquids X and Y boil at 383K and 403K respectively. Which one of them has higher
vapor pressure at 323K?
Ans: Liquid A has lower boiling pt. and hence is more volatile. Therefore, A has higher
vapor pressure at 323K.
6. What will happen when Red Blood Cell are placed in water?
Ans: Red blood cells will burst when placed in water
7. Of 0.1 molal solution of BaCl2and sodium chloride respectively, which one will have
a higher boiling point?
Ans: 0.1 molal solution of BaCl2 will have higher boiling point as BaCl2 disassociates in
the solution to give more no of ions.
8. What is the van’thoff factor for a compound which undergodimerisation in an organic solvent?
Ans: 2X
X2,
Hence van’thoff factor is ½.
9. 10 ml of liquid A was mixed with 10ml of liquid B. The volume of the resulting solution
was found to be 19.9ml. What do you conclude?
Ans. Negative deviation from Raoult’s law.
10. Plants grown in marshy lands generally show bursting of root hair. Why?
Ans: Due to osmosis water molecules comes out from the plants so root burst.
11. Why is the vapour pressure of a liquid constant at constant temperature?
Ans:- Because if it reaches a of equilibrium where rate of evaporation is the rate of condensation.
12. Define osmotic pressure.
Ans:- The minimum access pressure that has to be applied on the solution to prevent the entry
of solvent into the solution through SPM is called osmotic pressure.
13. Define ideal solution.
Ans. Ideal Solution is that which obeys raoult’s law at wide range of temperature
and concentration.
14. When is the value of van’t Hoff factor more than one ?
Ans. When the solute undergoes dissociation.
15. What is the value of van’t Hoff factor for K4[Fe(CN)6], if it is 100% dissociated?
Ans. 5
16.Define mole fraction.
Ans. Ratio of number of moles of a particular component to total number of moles of solution
17.Why are soda water and soft drink bottles sealed under high pressure?
Ans. Solubility increases.
18.How does osmotic pressure depend upon temperature ?
Ans. It increases with rise the temperature
19.Give one example each when van’t Hoff factor is 2 and 3
Ans. KCl and MgCl2
20. What will be the nature of the solution when ethyl alcohol and water are mixed ?
Ans. Positive deviation from Raoult s law
SA(2Mark)
1. Why does a solution of cyclohexane and ethanol show positive deviation from Raoult’s law?
Ans: On adding C6H12 to C2H5OH, its molecules get in between the molecules of ethanol
thus breaking the hydrogen bonds and reducing ethanol-ethanol interactions. This
will increase the vapor pressure of the solution and result in positive deviation from
Raoult’s law.
2. Why is liquid Ammonia bottle first cooled in ice before opening it?
Ans. At room temp., the vapor pressure of liquid Ammonia is very high. On cooling, the
vapor pressure is reduced, thus Ammonia will not splash out.
3. How does sprinkling of salt help in clearing the snow covered roads in hilly areas? Explain
the phenomenon involved in the process.
Ans: When salt is spread over snow covered roads, snow starts melting from the surface
because of the depression in freezing point of water and it helps in clearing the roads.
4. After removing the outer shell of two eggs in dil. HCl, one is placed in distilled water and
the other in a saturated solution of NaCl. What will you observe and why?
Ans: Eggs in water will swell whereas in NaCl solution it will shrink. This is because due
to osmosis, the net flow of solvent is from less concentrated to more concentrated solution.
5. Equimolal solutions of NaCl and BaCl2 are prepared in water. Freezing point of NaCl is
found to be -20C. What freezing point do you expect for BaCl2 solution?
Ans: For NaCl, van’thoff factor is 2, whereas for BaCl2, the van’thoff factor is 3. Hence
the depression of freezing point for BaCl2 solution is 3/2 that of NaCl solution. Hence
the freezing point of BaCl2 is -30C.
6. When fruits and vegetables are dried and placed in water, they slowly swell and return
to original form, why? Does an increase in temperature accelerate the process? Explain.
Ans. When fruits and vegetables are dried and placed in water, osmosis takes place, i.e.,
water molecules pass through semipermeable membrane present in cell membrane and
enter in to cell, therefore, they swell. If temperature is increased .osmosis will be
faster. Because osmotic pressure is directly proportional to temperature.
7. The depression in freezing point of water observed for the same amount of
acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given
above. Explain briefly.
Ans.The depression in freezing point of water observed for the same amount of
acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above.
It is because trifluoroacetic acid is stronger than trichloroacetic acid whereas acetic acid
is weakest acid.Stronger acid produces more number of ions, therefore has more T f ,
hence lower freezing point.
8. State Henry’s law and mention some important applications.
Ans: The pressure of gas over a solution is directly proportional to the mole fraction of the
gas dissolved in the solution.
Applications: (i) to dissolve more CO2 in cold drinks.
(ii)Deep sea divers use He and O2 mixture for respiration in place of N2 and O2 beacause
N2 being more soluble in blood at high pressure creates painful effect on human body.
9.(i) Name any two compounds which can be used as semi permeable membrane .
(ii)When two liquids A and B are mixed the solution becomes warmer,which deviation
solution exhibits from Raoult’s law?
Ans. (i)copper ferrocyanide and cellulose acetate.
(ii)negative deviation
10. Distinguish between the terms molality & molarity.
Ans.
Molality
Molarity
1
Number of moles of solute present per Number of moles of solute present per
kilogram of solvent
litre of solution
2
Does not depends on temperature
Depends on temperature
3
Mole/kg
Mole/litre
11. Define vapour pressure. What happens to the vapour pressure when
(i) A volatile solute is dissolved in the liquid?
(ii)A non-volatile solute is dissolved in the liquid?
Ans. The pressure exerted by gas molecules on its liquid layer at equilibrium.
(i)Increase
(ii)Decrease
12. 4% NaOH solution (mass/volume) and 6% urea solution (mass/volume) are equimolar
but not isotonic .Why?
Ans. NaOH undergoes dissociation in solution.
13.Calculate the mole fraction of benzene in a solution containing 30% by mass of it in CCl4
Ans. Mass of benzene =30g
Mass of CCl4 =70g
Molar mass of benzene= 78 g
Molar mass of CCl4= 154g mol-1
nC6H6=30g\78g=0.385mol
nCCl4=70g\154g=0.454mol
X C6H6 = n C6H6\n C6H6 + nCCl4 =0.385 mol \0.385mol + 0.454mol = 0.459
14.If the volume of a 100 L solution of 1.1 moles of hydrogen in 6.0 moles argon is
suddenly doubled, what happens to the mole fraction of hydrogen in that solution?
Ans. The mole fraction of a solute does not depend on the volume of the solution. Mole
fraction only depends on the molar amount of each component.
15. What is the concentration of HCl stock solution that can create 250 mL of a 1.0 M
solution that is prepared by the dilution of 50 mL of the HCl stock solution?
Ans. By rearranging the expression c1v1 = c2v2 to solve for c1, we find the concentration of
the HCl stock solution:
16. What is significance of Henry’s law constant KH ?
Ans. Henry’s law contant (KH) helps in comparing relative solubilities of different gases in
the same solvent.
17. Define ideal solution. Give one example.
Ans.A solution which strictly obeys raoult’s law at all concentrations and
temperatures. Example: benzene and toluene
18. Why is an increase of temperature observed on mixing chloroform with acetone.
Ans. Intermolecular H- bonding present in both components. Energy is needed to separate
them and this cause rise in boiling point of temperature.
19. State the condition leading to reverse osmosis.
Ans. Occurs when a pressure higher than osmotic pressure is applied.
20. What happens when RBC (RED BLOOD CELL ) is kept in brine and pure water?
Ans: In brine RBC shrinks due to exosmosis. In pure water it swells up due to endo osmosis
SA(3Mark)
1.Assuming complete ionization, calculate the expected freezing point of solution prepared
by dissolving 6.00g of Glauber’s salt, Na2SO4·10H2O in 0.1kg of H2O.
(Kf for H2O=1.86 K kg mol-1).
2. An aqueous solution of 3.12 g of BaCl2 into 250 g of water is found to boil at100.0832°C.
Calculate the degree of dissociation of BaCl2 .
Kb(H2O) = 0.52K/m.
3. Vapour pressure of CHCl3 and CH2Cl2 at 298K are 200 mm Hg and 415mm Hg respectively.
(i) Calculate the vapour pressure of the solution by mixing 25.5 g of CHCl3 &40 g of CH2Cl2
at 298K
(ii) Mole fraction of each component in vapour phase.
(Molar mass CHCl3 =118.5 CH2Cl2 = 85)
4. Define the term osmotic pressure.describe how the molecular mass of a substance can
be determined by the measurement of osmotic pressure?
Ans. Correct definition
MB=WBRT/V π
5. What do you mean by Raoult’s law ? What are the limitations of Raoult’s law ?
Ans. Raoult’s law – partial vapour pressure of any volatile component of a solution at
any tempreture is equal to the product of vapour pressure of the pure compenent and
mole fraction of the compoennt in solution give mathematical expression.
Limitations :
(i)Applicable only to dilute solutions.
(ii)Applicable to solutions of only non electrolytes
6.A solution containing 18 g of non-volatile solute in 200g of water freeges at 272.07 K.
calculate the molecular mass of solute (given Kf = 1.86 K/m)
Ans. W2 = 18 g W1= 200g, kf = 1.86k/m
Tf = 273K – 272.07K = 0.93K
1000×kf ×W2
1000×k ×W
Tf M ×W
 M2 = ∆M × fW 2
2
 M2 =
1
f
1000×1.86×18
1
= 180 amu
0.93×200
7. Calculate the temperature at which a solution containing 54 g of glucose (C6H12O6) in
250 g of water will freeze (kf for water =1.86 KKg/mol).
Ans. Given- W solute = 54 g , W solvent = 250 g, M solute = 180
Tf = kf x wsolute x 1000
Msolutex wsolvent
Tf = 1.86 x 54 x 1000
180x 250
Tf = 2.232 K
Tf = T0f - Tf
Tf = 273- 2.232
= 270.768 K
8. Define the term osmotic pressure.Describe how the molar mass of a substance can
be determined by a method based on measurement of osmotic pressure.
Ans. Pressure exerted to stop process of osmosis is called osmotic pressure.
∏ V =nRT
∏ = Osmotic Pressure
V = Volume of Solution [L]
n = No. of moles of solute
R = Gas Constant = 0.082 L atm / mol / K
T = Temperature [K]
W
M
=
RT
V
WRT
V
M = Molar Mass of Solute
W = Mass of Solute
M
W
V
=
M
W
V
=
s olute
=
d
dRT
=
M
=
=
C =
RT
n
V
RT
CRT
Mol ar Concentration [Molarity]
9. Calculate the mass percentage of benzene (C6H 6) and carbon tetrachloride (CCl4) if 22 g
of benzene is dissolved in 122 g of carbon tetrachloride.
Ans. Given- Mass of solute (WB ) = 22g
Mass of solvent (W B) = 122 g
Mass of solution = 22g + 122 g = 144 g
Now, Mass percentage of benzene
And, Mass percentage of carbon tetrachloride
Thus, Mass percentage of benzene = 15.27%
And, Mass percentage of carbon tetrachloride = 84.72%
10. Calculate the mole fraction of benzene in solution containing 30% by mass in
carbon tetrachloride.
Ans. Given, % of benzene by mass = 30%
Thus, % of solvent (carbon tetrachloride) = 70%
This means, 30g of benzene (C6 H 6) is dissolved in 70 g of CCl 4
Now, Molecular mass of benzene (C6 H6 ) = (12 × 6) + (1 × 6) = 78 g mol – 1
Molecular mass of carbon tetrachloride (CCl4 ) = 12 + (35.5 × 4 ) = 154 g mol – 1
Therefore, Number of moles of C 6H 6
And, number of moles of CCl4
Thus, mole fraction of benzene in the given solution
Thus, mole fraction of benzene in the given solution = 0.457
11. Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal
aqueous solution.
Ans. Given- Molality (m) = 0.25 m
Molar mass (M B) of urea (NH2CONH2) = 14 + 2 + 12 + 16 + 14 + 2 = 60 g mol – 1
LA(5Mark)
1. Why is the mass determined by measuring a colligative property in case of some solutes
abnormal ? Discuss it with the help of Van’t Hoff factor?
Ans: Certain compounds when dissolved in suitable solvents either dissociate or associate.
For example ethanoic acid dimerises in benzene due to hydrogen bonding, while in water, it
dissociates and forms ions. As a result the number of chemical species in solution increases or
decreases as compared to the number of chemical species of solute added to form the solution.
Since the magnitude of colligative property depends on the number of solute particles, it is
expected that the molar mass determined on the basis of colligative properties will be either
higher or lower than the expected value or the normal value and is called abnormal molar mass.
In order to account for the extent of dissociation or association of molecules in solution, Van’t
Hoff introduced a factor, i, known as the Van’t Hoff factor. It can be defined as follows.
i=Expected molar mass/Abnormalmolar mass
= Observed colligative property/Calculated colligative property
= (Total number of moles of particles after association/dissociation)/(Number of moles of particles
before association/dissociation)
2. (i)Why is the vapour pressure of an aqueous solution of glucose lower than that of water?
(ii) Why are aquatic species more comfortable in cold water in comparison to warm water?
(iii)Give one example of semi permeable membrane used in desalination of water .
(iv) why it is used?
Ans.(i) In pure liquid water the entire surface of liquid is occupied by the molecules of
water.When a non volatile solute, for example glucose is dissolved in water, the
fraction of surface covered by the solvent molecules gets reduced because
some positions are occupied by glucose molecules. As a result number of
solvent molecules escaping from the surface also gets reduced, consequently the
vapour pressure of aqueous solution of glucose is reduced.
(ii) At a given pressure the solubility of oxygen in water increases with decrease
in temperature. Presence of more oxygen at lower temperature makes the
aquatic species more comfortable in cold water.
(iii) Cellulose acetate.
(iv) Because it is permeable to water but impermeable to impurities and ions in sea water.
3. (i) 200 cm3 of an aqueous solution of a protein contains 1.26 g of the protein. The
osmotic pressure of such a solution at 300 K is found to be 2.57 × 10-3 bar. Calculate
the molar mass of the protein.
(ii) 45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water. Calculate(a) the freezing point depression
(b) The freezing point of the solution.
Ans. (i)The various quantities known to us are as follows: Π = 2.57 × 10–3 bar,
V = 200 cm3 = 0.200 litre
T = 300 K
R = 0.083 L bar mol-1 K-1
M2 = 1.26 g × 0.083 L bar K−1 mol−1 × 300 K / 2.57×10−3 bar × 0.200 L = 61,022 g mol-1
(ii) Moles of ethylene glycol =45/62 g mol-1=0.73 mol
Mass of water in kg =600g/1000g kg-1 = 0.6 kg
Hence molality of ethylene glycol = 0.73mol/0.60kg = 1.2 mol kg –1
∆ Tf = 1.86 K kg mol–1 × 1.2 mol kg –1 = 2.2 K
Freezing point of the aqueous solution = 273.15 K – 2.2 K = 270.95 K
4.If N2 gas is bubbled through water at 293 K, how many millimoles of N2 gas
would dissolve in 1 litre of water. Assume that N2 exerts a partial pressure of
0.987 bar. Given that Henry’s law constant for N2 at 293 K is 76.48 kbar.
Ans.The solubility of gas is related to the mole fraction in aqueous solution.
The mole fraction of the gas in the solution is calculated by applying Henry’s law.
Thus:x(Nitrogen)=p(nitrogen)/KH=0.987bar/76,480bar=1.29×10-5
As 1 litre of water contains 55.5 mol of it, therefore if n represents number of moles of
N2 in solution, x(Nitrogen) = n mol/ n mol + 55.5 mol =n/55.5= 1.29 & times10-5
Thus n = 1.29 × 10–5 × 55.5 mol = 7.16 × 10–4 mol
=7.16×10−4 mol × 1000 mol/1 mol = 0.716 mmol
5. (i)Define Azeotropes and explain briefly minimum boiling azeotropes by
taking suitable example.
(ii) The vapour presure of pure liquid A nad B are 450mm and 700mm
of Hg respectively at 350K calculate the compostion of liquid mixture if
total vapour pressure is 600 mm of Hg. Also find composition of mixtures
in vapour phase.
Ans.(i) A solution at fixed composition, which continues to boil at constant
temperature azeotropes without change in comperation of solution and its vapour
is called an azeotropes with minimum VP have maximum boiling point when
liquids in solution form strong chemical bonds, their escape tenncies and hence
V.P. decreases than expected on basis of Raoult’s law.
(ii) PA° = 450 mm PB° = 700mm
PA = PA° XA = 450XA and PB = PA° XB = 700XB
PA = PA° XA = 450XA + 700XB = 600
450XA + 700(1-XA) = 450XA + 700 -700XA = 600
250XA = 600-700 = -100
−100
2
XA = −250 = 5 = 0.4
XB = 1-4 = 0.6
PA = 0.4 x 450 = 180mm kg
PB = 0.6 x 700 = 420mm kg
PA
180
XA in vapour phase = P +P
= 180+420 = 0.3
A
B
XB in vapour phase = 1-0.3 = 0.7
6. Why it is advised to add ethylene glycol to water in car radiator while driving in a
hill station? 0.6 mL of acetic acid (CH3COOH), having density 1.06 g mL–1,
is dissolved in 1 litre of water. The depression in freezing point observed for
this strength of acid was 0.0205°C. Calculate the van’t Hoff factor and the
dissociation constant of acid.
Ans. Ethylene glycol lowers the freezing point of water and therefore water does
not In the radiator.
Acetic acid is a weak electrolyte and will dissociate into two ions: acetate and hydrogen ions per
molecule of acetic acid. If x is the degree of dissociation of acetic acid, then we would have
n(1 – x) moles of undissociated acetic acid, nxmoles of CH3COO– and nxmoles of H+ ions,
VALUE BASED QUESTIONS (4 MARKS)
1.The chairman, Kandla (Gujrat) port due to water scarcity has decided to desalinate sea water
to obtain potable water.
a. As a student of chemistry which method will be suitable to use? (1)
b. Discuss the method. (1)
c. What value have you inculcated in using this method?
Ans: a) Reverse osmosis.
b) Discussed
c) Using less energy ie energy conservation
2. Bharath went to his grandfather’s house in winter this year. As usual he
went for fishing. His grandmother told him there will be no fishes in the
lake. He noticed that it was more difficult to find fishes in winter. The
fishes were deep inside the river. Whereas in summer they were on the
surface and hence he was able to catch fishes.
a) Why are fishes on the surface in water than in the depth in summer?
(2)
b) What value can be derived from this? (1)
Ans:
a) According to Henry’s law at low temperature gases are more soluble
and hence as more oxygen gets dissolved in water fishes survive
better even in depth of the river. In summer as the oxygen is less in
water the fishes come to the surface.
c) The value that I derive from this is wisdom is superior to knowledge
3. Srisha wanted to keep ice creams without melting. So he had to keep it on
ice taken in a container. His grandmother advised him to pour salt on the ice.
a. Why? (2)
b. What is the value in this? (1)
Ans:
a. Adding salt to ice decreases the melting point of ice. Hence the
decrease in temperature of ice.
b. Obey your elders.
4. Sneha’s grandmother lives in Simla. In winter there is lot of snow in front of
the house. She asked Sneha to clear the snow from the front of the house.
Sneha added salt to snow to clear it.
a. Why?
b. What is the value in this?
Ans:
Adding salt to ice decreases the melting point of snow. The snow
melts. Hence snow can be removed.
b) Keep your surroundings clean
5.an untrained nurse used distilled water and not saline water while giving injection to the
patient. Immediate swelling was noticed in area on skin.
i) what mistake was committed by her?
ii) what precaution she must take in future?
IiI) which values should the nurse show in future?
Ans : !) distilled water is hypotonic as compared to blood cells. There is immediate flow of
excessive water into cell membrane and swelling occurs.
ii) she should use saline water which is isotonic with RBCs so that no swelling occurs.
iii) she should show carefulness and diligence while working.
Q33 Kalavati wanted to give her baby a medicine for fever. She added boiled and cooled water
as per the instructions, to the contents of the bottle, upto the mark. She shook the bottle. Then
gave a spoonful of the medicine to the baby. As a student of chemistry answer the following
questions-
Why did the shake up the contents? What is the process called?
What is value associated with selling medicine in this form?