RBSE Solutions for Class 12 Chemistry Chapter 2 Solution

RBSE Solutions for Class 12 Chemistry Chapter 2 Solution is part of RBSE Solutions for Class 12 Chemistry. Here we have given Rajasthan Board RBSE Class 12 Chemistry Chapter 2 Solution.

Rajasthan Board RBSE Class 12 Chemistry Chapter 2 Solution

RBSE Class 12 Chemistry Chapter 2 Text Book Questions

RBSE Class 12 Chemistry Chapter 2 Multiple Choice Questions

Question 1.
4 g of NaOH is dissolved in 500 g of water. The concentration of solution will be
(a) 8 g/L
(b) 0.2.N.
(C) 0.2 m
(d) 0.2 M

Question 2.
Which liquid pair represents positive deviation from Raoult’s law ?
(a) Water     + HCl
(b) Water     + HNO₃
(c) Benzene  + Methanol
(d) Acetone  + Chloroform

Question 3.
The molarity of pure water is-
(a) 55.5 M
(b) 100 M
(c) 18 M
(d) IM

Question 4.
Arrange the following 0.1 M solutions in increasing order of their boiling points
(a) NaCl
(b) MgCl₂
(c) Urea
(d) AICI₃
(a) (i) < (ii) < (iii) < (iv).
(b) (ii) < (i) < (iii) < (iv)
(c) (iii) < (i) < (ii) < (iv)
(d) (iv) < (iii) < (ii) < (i)

Question 5.
This is a property of an ideal solution
(a) It obeys Raoult’s law
(b) ∆H_mixing
(c) ∆V_mixing
(d) All of these

Question 6.
On increasing temperature, vapour pressure of a substance
(a) always increases.
(b) decreases.
(c) does not depend on temperature.
(d) partially depends on temperature.

Question7.
The osmotic pressure of 5% sugar solution will be
(a) 3.47 atm
(b) 5.07 atm
(c) 4.03 atm
(d) 2.09 atm

Question 8.
On increasing temperature, the solubility of H₂ gas in water
(a) increases
(b) decreases
(c) remains unchanged
(d) None of these
Answer:
1. (c), 2. (c), 3. (a), 4. (c), 5. (d), 6. (a), 7. (a), 8. (a)

RBSE Class 12 Chemistry Chapter 2 Very Short Answer Type Questions

Question 1.
Calculate molality of 10% (w/W) aqueous H₂SO₄
Answer:
Weight of solute (W_B) = 10 g
Weight of solvent (W_a) = 100 – 10 = 90 g
Molecular weight (M_B) = 98 g mol^-1

= 1.134 mol kg^-1

Question 2.
What is molarity. ? Write the effect of temperature on it.
Answer:
Molarity: It is the number of moles of solute dissolved per litre of the solution.

On increasing the temperature, molarity of solution decreases because on increasing temperature volume of solution increases .

Question 3.
Define the mole fraction of a substance in solution.
Answer:
Mole fraction-Mole fraction of a constituent (solute as well as solvent) is the fraction obtained by dividing number of moles of that constituent by the total number of moles of all the constituents present in the solution. In a binary mixture, there are two components A and B and their moles are n_A and n_B respectively. If their mole fractions are x_A and x_B

If the mass of A and B are W_A and W_B respectively and their molar masses are M_A and M_B then,
n_A = [latex]\frac { { W }_{ A } }{ { M }_{ A } } [/latex]
n_B = [latex]\frac { { W }_{ B } }{ { M }_{ B } } [/latex]

The sum of mole fractions of all the components are always equal to 1.
X_A + x_B = 1

Question 4.
Does it advise to use ethylene glycol in radiators of cars in summer ?
Answer:
No, it is not advised to use ethylene glycol in radiators of cars in summer because ethylene glycol decreases the freezing point of water, prevents the freezing of water of radiator in winter, so it is advised to use ethylene glycol in winter.

Question 5.
Define reverse osmosis ?
Answer:
If the pressure higher than the osmotic pressure is applied on the solution, the solvent will flow from the solution into the pure solvent through the semipermeable membrane. The process is called reverse osmosis.

RBSE Class 12 Chemistry Chapter 2 Short Answer Type Questions

Question 1.
Explain the effect of temperature on solubility of solid in liquid. Give the formula of van’t Hoff factor which relates the abnormal molecular mass to normal molecular mass. How does it affect by the association and dissociation?
Answer:
Effect of temperature on solubility of solid in liquid–There is following equilibrium between soluble solid and solution in a saturated solution. Insoluble solid + Solute – Solvent Solution ⇌ Solution ;
∆H_sol = ± x k cal
According to Le-chatelier’s principle if ∆H > 0 (zero) i.e. on dissolving solute in solvent, energy is absorbed, hence on increasing temperature the solubility of solid solute increases .
Example: NH₄ CI, KCI, AgNO₃, NaNO₃, KI etc.
If ∆H < 0 (zero) i.e. on dissolving solute in solvent energy is released, hence on increasing temperature the solubility of solid solute decreases.
Example: NaOH, Li₂SO₄, (CH₃ COO)₃ Ca etc.
van’t Hoff’s factor

If observed colligative property < calculated value or M_o > M_c (or i<1), there is association. If observed colligative property > calculated value or Mo < Mc (ori >1), there is dissociation.

Question 2.
The theoretical molecular mass and observed molecular mass of an ionic compound AB are 58.2 and 30 respectively. Calculate the van’t Hoff factor and degree of dissociation.
Answer:
Calculated molecular mass of AB = 58.2
Observed molecular mass of AB = 30

Degree of dissociation (α) = [latex]\frac { i-1 }{ n-1 }[/latex]
Substance AB on dissociation gives 2 mol substance A and B. Hence,

Question 3.
What are the differences between diffusion and osmosis ? Give an example of each. Represent the diffusion and osmosis by labelled digrams.
Answer:
Differences between diffusion and osmosis

Osmesis	Diffusion
1. In this process semi-permeable membrane is used.	1. In this process no semi-permeable membrane is used.
2. It takes place from lower concentration to higher concentration.	2. It takes place from higher concentration to lower concentration.
3. It takes place only in solutions.	3. It takes place in gases as well as in solutions.
4. It can be stopped or reversed by applying pressure on the solution  with higher concentration.	4. It cannot be stopped or reversed.

Examples:
Diffusion:
When potassium permanganate is added in container of water then after some time violet colour spreads in container. This is an example of diffusion.
Representation of process of diffusion by digram:

Osmosis:
If we put grapes in a concentrated solution of NaCl, the grapes shrink but if we put kismis or grapes into pure water, they swell. It is a phenomenon of osmosis.
Representation of process of osmosis by digram:

Question 4.
0.2L aqueous solution of protein contains 1.26 g protein. The osmotic pressure of this solution is 2. 57 x 10^-3 bar at 300 K. Calculate the molecular mass of protein.
Answer:
Given that,
Volume ‘V’ = 0.2 L
Mass of solute ‘W_B‘ = 1. 26 g
Temperature ‘T’ = 300 K,
Osmotic pressure ‘π’ = 2.57 x 10^-3 bar
Gas constant ‘R’ = 0. 0821 L bar mol^-1 K^-1


= 60377.04 g/mol
Molecular mass of protein = 60377.04 g/mol

Question 5.
Prove that ∆T_b = K_b.m for solution having non-volatile solute.
Answer:
If boiling point of solution is T_b and the boiling point of solvent is [latex]{ T }_{ b }^{ 0 }[/latex] then,
Elevation in boiling point [latex]\Delta { T }_{ b }={ T }_{ b }-{ T }_{ b }^{ 0 }[/latex]
Elevation in boiling point is directly proportional to lowering in osmotic pressure.
ΔΤ_b ∝ ΔΡ
According to Raoult’s law, lowering in vapour pressure is directly proportional to mole fraction of solute, hence,
ΔP α x_b (x₃ = mole fraction of solute)
Hence,
ΔΤ_b ∝ x_B

If W_A is the mass of solvent in kg, then [latex]\frac { { n }_{ B } }{ { W }_{ A } } [/latex] is equal to
molality (m) of the solution.
∆T_b = kM_Am
Here k and M_A are constants and hence their product
i.e., kM_A is replaced by another constant k_b.
∆T_b = K_bm
Where K_b is called boiling point-elevation constanr or molal elevation constant or molal ebullioscopic constant.

Question 6.
How can the molecular mass of non-volatile substance determined from the component of vapour pressure ? Explain it.
Answer:
For non-volatile substance, relative lowering in vapour pressure of solution is equal to the mole faction of solute.

Where,
n_B = Number of moles of solute
n_A = Number of moles of solvent
W_B = Mass of solute
W_A = Mass of solvent
M_B = Molecular mass of solute
M_A = Molecular mass of solvent
According to. Raoult’s law


With the help of above equation we can calculate the value of molecular mass of non volatile solute, if all other factors in equation are known.

Question 7.
What do you understand by solubility of gases? Explain the factors affecting solubility of gases in liquid.
Answer.
Solubility of gases in liquids: All gases are less or more soluble in water or soluble in certain limit and form a true solution. The volume (in mL) of a gas which dissolved in 1 mL of water is called absorption coefficient.The solubility of gases in liquids is affected by the following factors:

1. Nature of gas: The gases which react with solvent or ionized in solution are soluble in water. e.g. NH₃, HCl_(g) and SO₂ are more soluble in water and form NH₄OH, HCl_(l) and H₂SO₄ compounds respectively.
2. Nature of solvent: The polar gases are soluble in polar solvent while non-polar gases are soluble in non-polar solvents. It is like dissolves like rule e.g. HCl gas is more soluble in water than liquid benzene.
3. Effect of Temperature: Temperature has a marked effect on the solubility of a solid in liquid. The solubility may increase or decrease with the rise in temperature depending upon the value of ∆_sol H < 0. Let us recall, that saturated solution represents equilibrium between undissolved solute and dissolved solute.
   Solute + Solvent ⇌ Solution; ∆_sol H = ± x
   • If the value of ∆_sol H < 0 i.e. the solution process is exothermic, then according to Le-chatelier’s principle, the increase of temperature will push the solution equilibrium in the backward direction. In other words, the solubility of solutes decreases with rise in temperature. Some examples are Li₂SO₄, Na₂SO₄ etc.
   • If ∆_sol H > 0, i.e. solution process is endothermic, then increase of temperature will push solution equilibrium in the forward direction. In other words, the solubility of such solutes increases with the rise in temperature. Some examples of such solutes are KCI, KNO₃, NaNO₃ etc.
4. Effect of pressure: The solubility of gases is much affected by change in pressure. Henry proposed a relationship between pressure (P) and the composition of solution which is called Henry’s law. According to this law.
   “At constant temperature, the amount of gas soluble in unit volume of solvent is directly proportional to pressure created by gas at equilibrium on the surface of solvent”.
   If the mass of gas is m which is dissolved in unit volume 3 of solvent at pressure (P) then according to Henry’s law,
   m ∝ P
   m = kP
   k = Henry constant

Question 8.
Calculate the temperature at which the solution of 54g glucose present in 250g water will be freezed.
(K_f =1.86 K kg mol^-1)
Answer:
Given that:
Weight of solute (W_B) = 54g
Molecular weight of solute (M_B) = 180 g mol
Weight of solvent (W_A) = 250g
K_f = 1.86 K kg mol^-1
Depression in freezing point ∆T_f

Depression in freezing point = 2.23 K
Freezing point of solution = ?
Freezing point of water = 273.15 K
(∆T_f) = Freezing point of water – Freezing point of solution
Freezing point of solution = 273.15 – 2.23= 270.92

Question 9.
Draw a graph for an ideal solution made up by solute and solvent.
Answer: