L-2
SOLUTIONS
POINTS TO BE REMEMBERED:-
Solution is the homogeneous mixture of two or more substances.
The substances which make the solution are called components. Most of the
solutions are binary i.e., consists of two components out of which one is
solute and other is solvent.
Solute-
The component of solution which is present in smaller quantity.
Solvent–The
component of solution present in larger quantity or whose physical state is
same as the physical state of resulting solution.
Solubility-
The amount of solute which can be dissolved in 100 gm. of solvent at a
particular temperature to make saturated solution.
Solid solutions
are of 2 types -
1.
Substitutional solid solution e.g. Brass
(Components have almost similar size)
2. Interstitial
solid solution e.g. steel (smaller component occupies the interstitial voids)
Expression
of concentration of solution
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1.
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Mass
percentage= amount of solute in gm. present in 100
gm. solution.
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Percentage =
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Mass
of Solute (WB) x 100
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Mass of Solution
(WA+WB)
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For liquid
solutions percentage by volume is expressed as =
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Volume of Solute
(VB) x 100
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Volume of
Solution (VA+VB)
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2.
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Mole fraction: it is
the ratio of no. of one component to the total no. of moles of all
components. It is
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Expressed
as ‘x’. For two. component system made of A and B,
XA
= nA / nA+ nB
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XB=
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nB /
nA+ nB
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, Sum of mole
fractions of all the components is 1 ; XA+XB =1
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3. Molarity (M) = (No of moles of solute
/ Volume of Solution in litre)
It decreases with
increase in temperature as volume increases on increasing temperature
4. Molality (m)=(No
of moles of solute / Mass of Solvent in Kg.)
No effect of
change of temperature on molality.
5.Normality (N)
= (No of Gram equivalent of solute / Volume of Solution in litre)
It changes with
change in temperature.
Ppm= WB / (WA + WB)
x 106
Vapour pressure–It
is defined as the pressure exerted by the vapour of liquid over the liquid in equilibrium
with liquid at particular temperature. Vapour pressure of liquid depends upon
nature of liquid and temperature.
Raoult’s–
Law
1. For
the solution containing non-volatile solute the vapour pressure of the solution
is directly proportional to the mole
fraction of solvent at particular temperature
PA
α XA
PA
= P0A.XA
2. For
the solution consisting of two miscible and volatile liquids the partial vapour
pressure of each component is directly proportional to its own mole fraction in
the solution at particular temperature.
PA=P0A. XA, PB=P0B
.XB
And
total vapour pressure is equal to sum of partial pressure. Ptotal =
PA + PB
Ideal
solution –The solution which obeys Raoult’s law under all conditions of
concentration and during the preparation of which there is no change in
enthalpy and volume on mixing the
component.
Conditions
–
PA = P0A XA, PB = P0B.XB
This is only
possible if A-B interaction is same as A-A and B-B interaction.
Nearly ideal
solutions are –
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1.
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Benzene and
Toluene
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2. Chlorobenzene and Bromobenzene
3. n-Hexane and n-Heptane
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Non-ideal
solution –
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(a)
PA ≠P0A.XA
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(b)
PB ≠P0B.XB
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(b)
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∆H mix ≠0
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(d) ∆V mix ≠0
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For
non-ideal solution the A-B interaction is different from A-A and B-B
interactions
i.
For solution showing positive deviation
PA>
P0A XA& PB> P0B
XB
∆V Mix = positive, ∆Hmix=
positive (A-B interaction is weaker than A-A and B-B interaction)
e.g.
alcohol and water, ethanol and acetone, carbon- di- sulphide and acetone
ii.
For the solution showing negative
deviation
∆H Mix= negative, ∆V mix = negative
The
vapour pressures of two component systems as a function of composition (a) a
solution that shows positive deviation from Raoult's law and (b) a solution
that shows negative deviation from Raoult's law.
What is Azeotrope?
–The mixture of liquids at particular composition
which has constant boiling point which behaves like a pure liquid and cannot be
separated by simple distillation. Azeotropes are of two types:
(a)
minimum boiling Azeotrope (mixture which
shows +ve deviations ) ex. alcohol and water
(b)
maximum boiling Azeotrope (which shows –ve
deviations) ex. acetone and chloroform
Colligative Properties - Properties of ideal solution
which depends upon no. of particles of solute but independent of the nature of
particle are called colligative property.
Relative lowering
in vapour pressure:- (PoA
–PA)/ PoA = XB
Determination of
molar mass of solute:-
MB =( WB× MA× PoA)/WA
×(PoA –PA)
Elevation in
Boiling Point: ∆Tb = Kb. m
Where ∆T b
= Tb- Tob
Tb =
Boiling point of solution and Tob = Boiling point of
solvent
Kb = molal
elevation constant (Ebullioscopic constant)
m = molality
Depression in
Freezing Point:∆Tf =kf.m
Where∆T f =
T°f -Tf,
Tf =
Freezing point of solution and Tof
= Freezing point of pure solvent
m = molality Kf = molal depression
constant (Cryoscopic Constant)
unit = k.kg mol-1
Osmotic Pressure
The
hydrostatic pressure which is developed on solution side due movement of
solvent particles from lower concentration to higher concentration of solution
through semipermeable membrane. It is denoted as π and it is expressed as
π =CRT
π = (n/V) RT
n = No. of moles
of solute; v = volume of solution (L)
R = 0.0821 L atm K
-1mol-1; T = temperature in kelvin.
Isotonic solutions
have same osmotic pressure and same concentration.
Hypertonic
solutions have higher osmotic pressure and hypotonic solutions have lower
osmotic pressure.
0.91% solutions of
sodium chloride is hypertonic to blood so in this solution RBC would shrink and RBC swells up or burst in hypotonic
solutions.
(b) The henry’s law
constant for oxygen is 4.34×104atm at 25o
C. If the partial pressure of oxygen in air is 0.2 atm, under atmospheric
pressure conditions. Calculate the concentration in moles per Litre of
dissolved oxygen in water in equilibrium with water air at 25o C.
Ans:
Partial pressure of the gas is directly proportional to its mole fraction in
solution at particular temperature.PA αX A
PA
= KH XA,
KH
= Henry’s constant
KH
= 4.34×104atm
partial pressure
of oxygen(PO2)= 0.2 atm
SO as per henry’s
law PO2 =KHXo2
Xo2
= PO2 / KH =0.2
/ 4.34×104= 4.6×10-6
If
we assume 1L solution = 1L water
nwater
= 1000/18 = 55.5
XO2 =nO2/(nO2+ n H2O
) =nO2 /nH2O
nO2
= 4.6 X 10-6 X 55.5 = 2.55 X 10-4 mol
M
= 2.55 X 10-4 M
Q.2. What is Vant
Hoff factor?
Ans. It is the
ratio of normal molecular mass to observed molecular mass. It is denoted as ‘i’
i
= normal molecular mass / observed molecular mass
OR
I = no. of particles after association or
dissociation / no. of particles before association or dissociation
Q.3. What is the
Vant Hoff factor in K4[Fe(CN)6] and BaCl2 assuming
complete dissociation?
Ans 5 and 3 respectively
Q.4. Why the
molecular mass becomes abnormal?
Ans. Due to
association or dissociation of solute in given solvent .
Q.5. Define molarity,
how it is related with normality ?
Ans. N = M x
Basicity or acidity.
Ans. M = P x d
x10/Molar mass
Q.7. What role does the molecular interaction play in
the solution of alcohol and water?
Ans.
Positive deviation from ideal behavior .
Q.8. HowVant Hoff
factor is it related with
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a. degree of
dissociation
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b.degree of association
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Ans. a. α=(i –1)
/(n-1)
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b. α= ( i -1)
/1/n-1)
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Q.9. Why NaCl is
used to clear snow from roads?
Ans. It lowers
down the freezing point of snow
Q10. Why the
boiling point of solution is higher than pure liquid?
Ans. Due to
lowering in vapour pressure
HOTS
Q1. Out of 1M and 1m aqueous solution which is more
concentrated
Ans. 1M as density of
water is 1gm/ml
Q2. Henry law constant
for two gases are 21.5 and 49.5 atm, which gas is more soluble.
Ans. KH
is inversely proportional to solubility.
Q.3. Define
azeotrope , give an example of maximum boiling azeotrope.
(a)
Ans: The mixture of liquids at particular
composition which has constant boiling point which behaves like a pure liquid
and cannot be separated by simple distillation .Eg. acetone and chloroform
Q.4.
Calculate the volume of 75% of H2SO4 by weight (d=1.8
gm/ml) required to prepare 1L of 0.2M solution
Hint: M1
= P x d x 10 /98
M1
V1 = M2V2
14.5ml
Q.5.
Why water cannot be completely separated from aqueous solution of ethyl
alcohol?
Ans.
Due to formation of Azeotrope at (95.4%)
Q.1.
How many grams of KCl should be added to 1kg of water to lower its freezing
point to -8.00C (kf = 1.86 K kg /mol)
Ans. Since KCl
dissociate in water completely i=2
m=
8 / 2x1.86 = 2.15mol/kg.
Grams
of KCl= 2.15 x 74.5 = 160.2 gm.
Q.2. Define the following terms :
(i) Mole fraction
(x) (ii) Molality
of a solution (m)
Ans(i)it is the ratio of no.
of one component to the total no. of moles of all components.
(ii) It is number of
moles of solute per kilogram of the solvent.
Q.3. what do you
mean by colligative properties, which colligative property is used to determine
molecular mass of polymer and why?
Ans: Properties of ideal
solution which depends upon no. of particles of solute but independent of the
nature of particle are called colligative property. Osmotic pressure.
Q.4. Define
reverse osmosis, write its one use.
Ans. When the pressure more than osmotic pressure is
applied on the solution side, and the process of osmosis is reversed this is
called reverse osmosis. e.g. Desalination of water.
Q.5. Why does an
azeotropic mixture distills without any change in composition.
Hint: It has same
composition of components in liquid and vapour phase.
Q.6. Under what condition
Vant Hoff’s factor is
a. equal to 1 b. less than 1 c. more than 1
Ans: a) when no
association or dissociation of solute take place
a) in
case of association
b)
in case of dissociation
Q.7. If the density of some lake water is 1.25 gm /ml
and contains 92gm of Na+ ions per kg of water. Calculate the
molality of Na+ ion in the lake.
Ans. n = 92/23 =
4 m= 4/1 = 4m
Q.8. An aqueous solution of 2% non-volatile exerts a
pressure of 1.004 Bar at the normal boiling point of the solvent. What is the
molar mass of the solute.
Hint: P0A
–PA/P0A = wB X mA / mB
X wA
1.013
–1.004 / 1.013 = 2X 18 /mB X 98 ,
mB = 41.35gm/mol
Q.9. Why is it
advised to add ethylene glycol to water in a car radiator in hill station?
Because it make
the anti-freezing solution and lower the freezing point of water.
Q.10. what do you mean by hypertonic solution, what
happens when RBC is kept in 0.91% solution of sodium chloride?
Ans: Hypertonic solutions have higher osmotic
pressure. 0.91% solutions of sodium chloride is hypertonic to blood so in this
solution RBC would shrink
SHORT
ANSWER QUESTIONS (3 MARKS)
Q1.A solution containing 15 g urea (molar
mass = 60 g mol–1) per litre of solution in water has the same
osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol–l)
in water. Calculate the mass of glucose present in one litre of its solution.
Q2. A solution of glucose (molar mass =
108 g mol–1) 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–1)
Q3.1.0 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. (Kffor benzene = 5.12 kg mol–1)
Q4. 15.0 g of an unknown molecular
material is dissolved in 450 g of water. The resulting solution freezes at – 0.34°C. What is the molar
mass of the material? (Kffor water = 1.86 K kg mol–1)
Q5.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 100.42°C. What mass of glycerol was dissolved
to make this solution? (Kb for water = 0.512 K kg mol–1)
Q6.A solution prepared by dissolving 8.95
mg of a gene fragment in 35.0 mL of water has an osmotic pressure of 0.335 torr
at 25°C. Assuming that the gene fragment is a non-electrolyte, calculate its
molar mass.
Q7.What mass of NaCl must be dissolved in
65.0 g of water to lower the freezing point of water by 7.50°C? The freezing
point depression constant (Kf) for water is 1.86 C/m. Assume van’t Hoff
factor for NaCl is 1.87. (Molar mass of NaCl = 58.5 g).
Q8.100 mg of a protein is dissolved in
enough water to make 100 mL of a solution. If this solution has an osmotic
pressure 13.3 mm Hg at 25° C, what is the molar mass of protein? (R =
0.0821 L atmmol–1K–1 and 760 mm Hg = 1 atm.)
Q9. Calculate the amount of KCl which must
be added to 1 kg of water so that its freezing point is depressed by 2 K. (Kffor
water =1.86 K kg mol-1, Atomic mass
(K
= 39, Cl = 35.5)
Q10.What concentration of nitrogen should
be present in a glass of water at room temperature? Assume a temperature of 25°
C, total pressure of 1 atmosphere and mole fraction of nitrogen in air of 0.78.
[KH for nitrogen = 8.42 × 10–7 M/mm Hg]
Value
based questions
Q.1 Scuba divers when come towards the surface, the pressure gradually
decreases resulting in the released of dissolved gases leading to formation of bubbles of nitrogen gas in
the blood which blocks the capillaries and thus harmful kinds are created. To
avoid bends and toxic effects of high concn of nitrogen gas, the air is diluted
with helium. After reading the above
passage, answer the following questions
i) Why is the harmful condition of bends overcome by the use of helium?
ii) Which law is used to calculate the concentration of gases in solution?
i)
Mention the value
associated with providing divers air diluted with helium.
ii)
Q.2 Ram takes a open pan to cook vegetables at a hill station while
shyam cook the same vegetables in a pressure cooker at the same place
(a) Explain with reason who will cook vegetable faster.
(b) Mention the reason for the delay in cooking.
(c) Which value is learnt by the student in the process of cooking food
in the pressure cooker?
Q.3
a) Name the process observed when pressure on solution
side is more than osmotic pressure
(b) Write main use of this process.
(c) Mention the values associated with the above process
Q.4Bharath 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. hereas 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?
b) What value can be derived from this?
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.
b) The value that
I derive from this is wisdom is superior to knowledge
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