2 or 3 marks CHSE/CBSE Chemistry Questions and Solutions
(Topic: Solution)
1. State and explain Henry's law.
Ans: The dependence of solubility of a gas in a solvent on pressure can be quantitatively expressed by Henry’s Law. This law can be expressed in a number of ways:
(a) The solubility of a gas (in mol ltr-1) in a liquid at a particular temperature is directly proportional the partial pressure of the gas in equilibrium with the liquid at that temperature.
Mathematically, s α p => s = k p, where s= solubility, Unit of k = mol L-1 atm-1
(b) The amount of a gas dissolved (expressed in terms of mole fraction) in a definite volume of solvent at constant temperature is directly proportional to the partial pressure of the gas present in equilibrium with the solvent at that temperature.
Mathematically, pg = kH Xg
Where, KH is the Henry law constant. A gas having high solubity comparision to other has lower KH value.
2. What is relative lowering of vapour pressure?
Ans: The lowering in vapour pressure of a solution w.r.t. the vapour pressure of the pure solvent is called the rlative lowering in vapour pressure.
The difference between the vapour pressures of a pure solvent and the corresponding solution is called the lowering in vapour pressure. When this lowering in vapour pressure is compared to the vapour prssure of the pure solvent, it is called teh relative loweriing in vapour pressure.
Mathematically, (p0A - ps) / p0 = Xsolute. Where p0A - ps is the lowering in vapor pressure, and thus (p0A - ps) / p0 is called the relative lowering in vapor pressure which is equal to the mole fraction of the solute.
3. What is the reason of elevation in boiling
point?
Ans: When a nonvolatile solute is added to
a pure solvent, the vapor pressure of solution thus formed becomes less than
that of the pure solvent. Now the solution should be heated more in comparison
to the pure solvent in order to get enough vapors whose pressure becomes equal
to the atmospheric pressure in order to attain the boiling point. Therefore a
solution always boils at a higher temperature than the pure solvent.
The increase in the boiling temperature of a solution w.r.t. to the
boiling point of the corresponding pure solvent is called the elevation in
boiling point.
4. State Roult's law for solution containing
volatile solute.
Ans: In Raoult’s law, the relationship
between the vapor pressure of the solution and the corresponding pure solvent
and the amount of solute dissolved in the solvent has been established. This
can be discussed under two subsections:
a. Solution containing
volatile solute: The vapor pressure of a component in
a volatile mixture is directly proportional to the mole fraction of that
component.
Mathematically, pA α
XA and pB α XB => pA =
p0A XA and pB =
p0B XB , where pA and
pB are the vapor pressures due to component A and B in
the volatile mixture. XA and Xb are the mole
fractions of the component A and B. p0A and p0B are
the vapor pressure of pure solvent A and B.
b. Solution containing
nonvolatile solute: The vapor pressure of a solution is
directly proportional to the mole fraction of the solvent.
Mathematically, ps α XA =>
ps = p0A XA , where ps = v.p of the solution and XA is the mole fractions of the solvent.
p0A is the vapor pressure of pure
solvent.
5. Draw the Graph of elevation in boiling
point.
Ans:
Where, P0 = V.P of pure solvent, T0 = B.P of pure solvent and T1 and T2 are the boling temperatures of solution 1 and solution 2.
T1 - T0 = Elevation in boling point of solution 1
T2 - T0 = Elevation in boling point of solution 2
6. Draw the graph of depression in freezing
point.
Ans:
Where, P0 = V.P of pure solvent, T0 = F.P of pure solvent and T1 and
T2 are the freezing temperatures of solution 1 and solution 2.
T0 - T1 = Depression in Freezing point of solution 1
T0 - T2 = Depression in Freezing point of solution 2
7. What is molal elevation constant?
Ans: Molal elevation constant or Ebbulioscopic constant is
defined as the elevation in boiling point of one molal solution.
We
know, ΔTb = kb m’
. Thus when molality = 1 m, ΔTb = kb
.
Accordingly we
wrote the definition in that way.
The unit of
molal elevation coinstant is kelvin kg per mole and this constant for a
particular solvent. The molal elevation constant of water is 0.52 kelvin
kg per mole.
8. What is molal depression constant?
Ans: Write similarly
9. State Avogadro vant Hoff law of osmotic
pressure.
Ans: Equal volume of all solutions having the same osmotic pressure
and temperature contain the same number of solute molecules or particles.
For example, 1 litre solution of 1 M
urea and 1 litre of M/2 NaCl solution exert same osmotic pressure at same
temperature, as they contain equal number of particles.
Solutions having same osmotic
pressure at same temperature are called isotonic or iso-osmotic solution.
10. How can you determine the molecular weight
of an unknown solute from osmotic pressure of a solution.
Ans: By developing the van't Hoff's ideal equation for dilute
solution further we can have a formula from which we can dtermine the molecular
mass of an unknown solute:
πV = nRT, this is the van’t Hoff equation for dilute solution.
Again n, the number of moles = w/m (given mass of
solute/molecular mass of solute)
=> πV = (w/m)RT, knowing all other
values we can calculate the molecular mass of an unknown solute using this
formula.
11. What is the relationship between vapour
pressure s of volatile component and molefeaction of them in their solution?
Ans:
Roult's law for solution containing volatile solution is: The vapor pressure of a component in a volatile mixture is directly proportional to the mole fraction of that component.
Mathematically, pA α XA and pB α XB => pA = p0A XA and pB = p0B XB , where pA and pB are the vapor pressures due to component A and B in the volatile mixture. XA and Xb are the mole fractions of the component A and B. p0A and p0B are the vapor pressure of pure solvent A and B.
The vapour pressure of the
solution will be equal to,
ps = pA +
pB = p0A
XA + p0B XB
= p0A XA + p0B
(1 – XA) = (p0A - p0B)
XA + p0B
12. State Boyles vant Hoff law.
Ans: Write the statement and the mathematical formula.
13. State Charle's vant Hoff law.
Ans: Write the statement and the mathematical formula.
14. What is binary solution and what are its
type?
Ans: A solution containing two components
is called a binary solution.
Depending on the physical states of the solvents, binary solutions can be classified into the following types:
1. Liquid Solution: Here the solvent is a liquid and solute can be either solid or liquid or gas. For example: Sugar in water, ethanol in water and oxygen in water.
2. Gaseous Solution: Here the solvent is a gas and solute can be either solid or liquid or gas. For Example: Camphor or chloroform or oxgen in nitrogen.
3. Solid Solution: Here the solvent is a solid and solute can be either solid or liquid or gas. For Example: Copper in gold, mercury in sodium (sodium amalgam) and hydrogen in paladium.
15. What is solid solution and what are its
type?
Ans:
Solid Solutions: The solution in
which both the solute and solvent are solids is called a solid solution. There
are two types of solid solutions:
(1) In Substitutional Solid Solution, the solute
particles occupy (take the place of) the position of solvent particles in a
crystal lattice. In this case the atomic sizes of the solute and solvent
particles are equivalent. Example: Brass, Bronze, Alloys of Copper and
nickel, Silver and gold, Molybdenum and tungsten, etc.
(2) In Interstitial
Solid Solution, the solute particles occupy (take the place of) the
interstitial positions in the crystal lattice. These solutions are formed because
the solute atoms are small enough to fit into the interstitial positions
between the solvent atoms. Example: Carbon atoms occupying the interstitial
positions among iron solvents atoms in steel, hydrogen atoms occupying the
interstitial positions among palladium or platinum solvent atoms, etc.
16. 3g urea is dissolved in hundred ml water.
find the freezing point of solution (kf = 1.86k kg per mole)
Ans:
Since density of water = 1g/ml, mass of solvent = W =100g,
weight of solute = w = 3g, freezing point of pur solvent = 0 deg centigrade =
273k, m = molecular mass of solute (urea) = 60,
ΔTf = (kf . w . 1000) / (m . W), Thus calculate ΔTf and subtract it from 273k to get the answer
17. What are the conditions to behave like
Ideal solutions. Give examples.
(This question carries in actual 3 or 5 marks)
Ans: The solutions which obey Raoult’s law
and satisfy the following conditions are called ideal solutions.
Condition 1: Change in volume before and
after mixing the solvents (to prepare the solution) is equal to zero.
i.e., ΔVmix = 0
Conditions 2: Change in enthalpy before and after
mixing the solvents (to prepare the solution) is equal to zero. i.e., ΔHmix =
0.
From Raoult's Law, pA = p0A
XA and pB = p0B
XB , where pA and pB are the
vapor pressures due to component A and B in the volatile mixture. XA and
XB are the mole fractions of the component A and B. p0A and
p0B are the vapor pressure of pure solvent A and B.
Thus, P(total)
= PA+PB = P0A
XA + P0B XB it’s
called an ideal solution (graph:1 below). This happens when force of attraction
between component A and B (FA-B) is equal to the force of attraction between component A and A
(FA-A) and
the force of attraction between component B and B (FB-B). i.e., FA-B = FA-A = FB-B. Clearly very dilute
solutions prepared from liquids having similar nature and structures are
ideal. Example: 1. Mixture of methanol and ethanol 2. Mixture
of n-hexane and n-heptane 3. Mixture of benzene and toluene.
18. What is osmosis and what are its type?
Ans: The
movement of solvent molecules from less concentrated solution to the higher one
when both the solutions are separated by a semepermeable membrane is called the
osmosis.
When solvent molecules pass out of the cell due to osmosis, it is called exo-osmosis and when passe into the cell, it is known as endo-osmosis. A peeled egg shrinks when placed inside concentrated NaCl solution. This is becase water content from inside the egg comes out off egg. This is the example of exoosmosis. a peeled egg swells when placed in pure water because water enters into the egg due to concentration difference. this is called endoosmosis.
19. What is Freezing point? Draw a graph for
the depression in freezing point of a
solution.
Ans: The
temperature at which both solid and liquid co exist and both have the
same
vapour
pressure is called the freezing point.
Here,
T0 = freezinf point of pure solvent,
T1 =
Freezing point point of solution 1
T2
= Freezing point point of solution 2
20.
What is elevation in boiling point? Draw a graph for the elevation in
boiling
point
of a solution.
Ans: Give definition and draw the graph.
21. Draw the graph for ideal solution.
22.
Draw the graph for non ideal solution showing positive and negative deviation.
Ans:
Here ΔVmix > 0 , ΔHmix > 0
and P1 + P2 > Ptotal, where ΔVmix, ΔHmix, P1,
P2 and
23.
What is azeotrope? Give example.
Ans: Ans: A mixture of two liquids which has a constant boiling point and composition throughout the distillation is called the Azeotrope. In other words, mixtue of liquids which boils at constant temperature and have same composition both in solid and liquid phase is called an azeotrope.
If the mixture shows positive
deviation from Raoult's law then the azetrope will boil at a temperature
which is less than the boiling point of any of the component. This type of
azeotrope is called positive azeotrope or minimum boiling azeotrope.
On the other hand if the
mixture shows negative deviation from Raoult's law then the azetrope
will boil at a temperature which is higher than the boiling point of any
of the component. This type of azeotrope is called negative azeotrope
or maximum boiling azeotrope.
Example: The ethanol water mixture (95.63%
C2H5OH + 4.37 % by mass H2O) is a minimum boiling azeotrope as it
boils at 78.2 degree centigrade which is less than the boiling point of
ethanol (78.4 unit) and water (100 unit)
24. Explain why freezing point depression of
0.1 M NaCl is nearly twice that of 0.1 M glucose solution.
Ans: Glucose is a nonelectrolyte where as NaCl is strong electrolyte which dissociates completely. The number of particles formed in 0.1M NaCl is almost double of that formed in 0.1M glucose solution. Since colligative properties are directly proportional to the number of particles. Thus double number of particle in 0.1M NaCl is the reason why its F.P is double of that glucose.
25. What is osmotic pressure? Why is it a
colligative property?
Ans: It can be defined as the excess
pressure which must be exerted on the solution side (more
concentrated) in order to check the flow of solvent molecules towards
the solution side (more concentrated) when two solutions are
separated by a semi permeable membrane.
According to B.V
law, π α C, at constant temperature
According to C.V law, π
α T, at constant concentration
On combination of the
above two laws,
π α CT => π = CRT , where π, C, R and T are osmotic pressure, concentration, universal gas constant and temperature respectively.
Since concentration = n/v => π =
(n/V) RT => πV = nRT, where n and v are
number of moles and volume respectively.
Again n = w/m (given
mass of solute/molecular mass of solute)
=> πV = (w/m)RT,
Since osmotic pressure depends on the
molecular mass of the solute, its a colligative property.
26. What weight of the non volatile urea must
be dissolved in hundred gram of water in order to decrease its vapour
pressure by 25%?
Ans: Let V.P of pure solvent = 100, Thus
V.P of Solution = 100 - 25 = 75%
As we know, relative lowering in vapour
pressure = (P0 - Ps) / P0 = wM / Wm
W = 100 gram, m = molecular mass of urea = 60,
M = 18 and w = ?
Thus (100 - 75) / 100 = w . 18 / 100 . 60
Therefore, w =weight of urea = 83.3
g
27. What is van't Hoff factor? What is its value for the dissociation and association of the solute in solution?
Ans: The values of colligative properties are directly proportional
to the number of particles in the solution. The actual number of particles
in the solution change when association or dissociation of solute
particles takes place. Accordingly, the molar mass of solute calculated in
this way deviate from true molar mass hence called abnormal molar mass.
The van't Hoff factor ' i ' relates to the abnormal
colligative properties hence abnormal molar mass.
It is defined as the ratio of
number of particles after association or dissociation to the actual number
of solute particles taken.
i = number of particles after association or dissociation / actual number of solute particles taken.
i = Experimental colligative property / calculated colligative property
i = calculated molecular mass / experimental molecular mass
The number of particles after association is
less than the actual particles taken hence i < 1
similarly in case of dissociation i > 1.
28. The osmotic pressure of a solution is 2.4
atm at 27 degree centigrade if its strength is 18 gram per litre find
molecular mass of the solute.
Ans: π = 2.4 atm, T = 300k, w = 18 g, v = 1 litre, m = ?
π = wrt / mv, now put the values and get the
answer.
29. Calculate the molality of KCl solution in
water if its freezing point is 271 Kelvin.
Ans: T0 = 273k, T' = 271k, ΔTf = 273 - 271 = 2K, KCl gives 2
particles on dissociation,
so i = 2, Kf = 1.86
ΔTf = i Kf m' ....... get the
answer
No. 30 to 35 is for you. Solve yourself.
30. How many grams of glucose must be dissolved in 0.5 litre of water for its osmotic pressure to be the same as that of a solution of 12 gram urea in 250 ml water.
31. If mole fraction of a solvent is 0.25 and its vapour pressure in pure state is 150 mm find the vapour pressure of the solution.
32. 10 gram of non volatile solute was
dissolved in hundred gram water at 20 degree centigrade. The vapour
pressure was lowered from 17.35 to 17.2 3mm calculate molecular mass of
the solute.
33. A 5% solution of sucrose is isotonic with 3% aq. solution of another substance A. Find molecular mass of A.
34. Dissolution of 10 gram solute in hundred gram of benzene increases its boiling point by 1 degree centigrade calculate molecular mass of solute (kb = 2.35 Kelvin kg per mole)
35. How much urea must be added to 500 gram
water so that it will boil at 374 Kelvin. (Kb = 0.52 Kelvin kg per
mole)