CBSE Class 12 Chemistry Notes Chapter 2 Solutions

Table of Contents

Topics in CBSE Solutions for Class 12 Chemistry Chapter 2 Solutions:

Topic Covered
Solutions
Types of Solutions
Expressing Concentration of Solutions
Solubility
Vapour Pressure of Liquid Solutions
Ideal and Non-ideal Solutions
Colligative Properties and Determination of Molar Mass
Abnormal Molar Masses

CBSE or NCERT Textbook Question Solved

2.1 Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride

Ans: Mass % of benzene

Mass percentage of C6H6=(Mass of C6H6)/(Mass of solution)×100 = 22×100/22+122=15.28%

Mass percentage of ℂl4 =(Mass of CCL4)/(Mass of solution)×100 = 122×100/22+122=84.72%

2.2 Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride

Ans: Let the total mass of the solution be 100 g and the mass of benzene be 30 g.

∴ Mass of carbon tetrachloride = (100 – 30) g = 70 g

Molar mass of benzene (C6H6) = (6 × 12 + 6 × 1) g mol – 1= 78 g mol – 1

∴ Number of moles of C6H6 =30/78 mol= 0.3846 mol

Molar mass of carbon tetrachloride (CCl4) = 1 × 12 + 4 × 35.5= 154 g mol – 1

∴ Number of moles of CCl4 = 70/154 mol= 0.4545 mol

Thus, the mole fraction of C6H6 is given as:

= 0.3846 / (0.3846 + 0.4545)= 0.458

2.3 Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO3)2. 6H2O in 4.3 L of solution
(b) 30 mL of 0.5 M H2SO4 diluted to 500 mL.

Ans: Molarity is given by

(A) Given : Co(NO3 )2. 6H2O = 30 gm
Volume of solution = 4.3 liter
We know :
Molar mass of Co(NO3 )2. 6H2O = 59 + 2(14 + 3 × 16) + 6 × 18 = 291 g mol−1
Moles of (NO3 )2. 6H2O = (weight)(gm)
molecularweight = 30 291 mol = 0.103 mol
The formula used : Molarity = Moles of solute Volume of solution in litre
Therefore, molarity= 0.103 mol 4.3 L = 0.023 M

(B) Given : 30 mL of 0.5 M H2SO4
Initial Volume of H2SO4 solution =30 ml
Number of moles present in 1000 mL of 0.5 M H2SO4 = 0.5 mol
Number of moles present in 30 mL of 0.5 M H2SO4
= 0.5×30 1000 mol = 0.015 mol
Therefore,Molarity = Moles of solute/ Volume of solution in litre
= Moles of solute×1000 /Volume of solution in Mili litre
Moles of solute = MV(litre)
Final Volume of the solution after dilution = molarity = MiVi/ Vf
= 0.015/0.5 L mol=0.03 M

2.4 Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution. 2.5 Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1

Ans: Molar mass of urea (NH2CONH2)= 2 (1×14+2×1)+1×12+1×16= 60g mol – 1
0.25 molar aqueous solution of urea means:
1000 g of water contains 0.25 mol= (0.25 × 60) g of urea = 15 g of urea
That is, (1000 + 15) g of solution contains 15 g of urea
Therefore, 2.5 kg (2500 g) of solution contains = (15 X 2500) / (1000+15) g= 36.95 g= 37 g of urea (approximately)
Hence, mass of urea required= 37 g

2.5 Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1

Ans: (A) Molar mass of KI = 39 + 127 = 166 g mol – 1
20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution.
That is,
20 g of KI is present in (100 – 20) g of water = 80 g of water
Therefore, molality of the solution = Moles of KI / Mass of water in kg
= 20/166 / 0.08 m
= 1.506 m
= 1.51 m (approximately)

(B) It is given that the density of the solution = 1.202 g mL – 1
∴Volume of 100 g solution = Mass /  Density
= 100g / 1.202g mL-1
= 83.19 mL
= 83.19 × 10 – 3 L
Therefore, molarity of the solution = 20/166 mol / 83.19 × 10 – 3 L= 1.45 M

(C) Moles of KI = 20/166 = 0.12 mol
Moles of water = 80/18 = 4.44 mol
Therefore, mole fraction of KI = Moles of KI / (Moles of KI + Moles of water)
= 0.12 / (0.12+4.44)= 0.0263

2.6 H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.195 m, calculate Henry’s law constant

Ans: 1kg of solvent contains 181000​=55.55 moles.
Mole fraction of H2​S is 0.195+55.550.195​=55.4750.195​=0.0035
Pressure of H2​S  =0.03550.987​=282bar
( at STP pressure is 0.987bar.)

2.7 Henry’s law constant for CO2 in water is 1.67×108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K.

Ans: According to Henry’s law, pCO2 = KHx
Here x is the mole fraction of carbon dioxide in the carbon dioxide water mixture.
pCO2 = 2.5 atm = 2.5 * 1.01325 * 105 Pa = 2.533125 * 105 Pa
x = mole fraction = 2.533125 * 105/ 1.67 * 108 = 0.00152
x = moles of CO2/(moles of CO2 + moles of H2O) ≈moles of CO2/moles of H2Oas mole of CO2are negligible in comparison to the number of moles of water
500 ml of water = 500g of water = 500/18 = 27.78 moles of water.
So, moles of CO2/27.78 = 0.00152 or moles of CO2 = 0.042 moles.
Hence the amount of CO2 in 500 ml of soda water is 0.042 * 44 = 1.848 g

2.8 The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K . Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

Ans: Let the composition of liquid A (mole fraction) be xA.
So mole fraction of B will be xB = 1 – xA.
Given that,    
Using Raoult’s law ,                  
Putting values of ptotal and vapour pressure of pure liquids in the above equation, we get :
                                    600 =   450.xA  +    700.(1 –  xA)
or                           600 – 700  =  450xA – 700xA
or                                  xA   =   0.4
and                               xB    =   0.6
Now pressure in vapour phase : 
                                        =   450(0.4)  = 180 mm of Hg
                                        =  700(0.6)   = 420 mm of Hg                                                                  And mole fraction of liquid B = 0.70

2.9 Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2 ) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Ans: Given, Vapour pressure of water, PIo = 23.8 mm of Hg
Weight of water, w1 = 850 g
Weight of urea, w2 = 50 g
Molecular weight of water, M1 = 18 g/mol
Molecular weight of urea, M2 = 60 g/mol

n1=w1/M1=850/18=47.22 mol

n2=w2/M2=50/60=0.83 mol

We have to calculate vapour pressure of water in the solution p1

By using Rault’s theorem,

P1 = 23.4 mm of Hg

Hence,

The vapour pressure of water in the solution is 23.4mm of Hg and its relative lowering is 0.0173.

2.10 Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.

Ans: Mass of water, w1 = 500 g
Boiling point of water = 99.63°C (at 750 mm Hg).
Molal elevation constant, Kb = 0.52 K kg/mol
Molar mass of sucrose (C12H22O11), M2 => (11 × 12 + 22 × 1 + 11 × 16) = 342 g/mol
Elevation of boiling point ΔTb = (100 + 273) – (99.63 + 273) = 0.37 K
We know that,

ΔTb = Kb*1000*w2 / M2*w1

0.37 = 0.52*1000*w2 / 342*500

w2 = 0.37*342*500 / 0.52*1000

w2 = 121.67 g

Hence, 121.67 g (approx) Sucrose is added to 500g of water so that it boils at 100°C.

2.11 Calculate the mass of ascorbic acid (Vitamin C, C6H8O6 ) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. Kf = 3.9 K kg mol-1 .

Ans: Mass of acetic acid, w1= 75 g Molar mass of ascorbic acid (C6​H8​O6​),
M2=6×12+8×1+6×16 =176gmol−1
Lowering of melting point, ΔTf​=1.5K
We know that:
ΔTb ​=M2​×w1​Kb​×1000×w2​​
w2 ​=Kb​×1000ΔTb​×M2​×w1​​
w2 ​=3.9×10001.5×176×75​
w2​ =5.08g
Hence, 5.08 g of ascorbic acid is needed to be dissolved.

2.12 Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.

Ans: Osmotic Pressure= π =(n/v)RT

We are given with: Moles of Polymer=1/185000

Volume, V = 0.45 L

Thus osmotic pressure= [ (1/185000) * 8.314 * 103 * 310 ] / 0.45 = 30.98Pa

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