NCERT Solution for Class 12 Chemistry Chapter 1: Solutions

In case of alcohol as well as water, the molecules are interlinked by intermolecular hydrogen bonding. However, the hydrogen bonding is also present in the molecules of alcohol and water in the solution but it is comparatively less than both alcohol and water. As a result, the magnitude of attractive forces tends to decrease and the solution shows positive deviation from Raoult’s Law. This will lead to increase in vapour pressure of the solution and also decrease in its boiling point.

Mass of solution = Mass of C6H6 + Mass of CCl4
= 22 g+122 g= 144 g
Mass % of benzene = 22/144 x 100 =15.28 %
Mass % of CCl4 = 122/144 x 100 = 84.72 %

0.25 Molal aqueous solution to urea means that
moles of urea = 0.25 mole
mass of solvent (NH2CONH2) = 60 g mol-1
.’. 0.25 mole of urea = 0.25 x 60=15g
Mass of solution = 1000+15 = 1015g = 1.015 kg
1.015 kg of urea solution contains 15g of urea
.’. 2.5 kg of solution contains urea =15/1.015 x 2.5 = 37 g

 When gases are dissolved in water, it is accompanied by a release of heat energy, i.e., process is exothermic. When the temperature is increased, according to Lechatlier’s Principle, the equilibrium shifts in backward direction, and thus gases becomes less soluble in liquids.

 Solubility of H2S gas = 0.195 m
= 0.195 mole in 1 kg of solvent
1 kg of solvent = 1000g

Vapour pressure of pure liquid A (

P

∘

A

) = 450 mm
Vapour pressure of pure liquid B (

P

∘

B

) = 700 mm
Total vapour pressure of the solution (P) = 600 mm

A solution is a homogeneous mixture of two or more chemically non-reacting substances. Types of solutions: There are nine types of solutions.
Types of Solution Examples
Gaseous solutions
(a) Gas in gas Air, mixture of 02 and N2, etc.
(b) Liquid in gas Water vapour
(c) Solid in gas Camphor vapours in N2 gas, smoke etc.
Liquid solutions
(a) Gas in liquid C02 dissolved in water (aerated water), and 02 dissolved in water, etc.
(b) Liquid in liquid Ethanol dissolved in water, etc.
(c) Solid in liquid Sugar dissolved in water, saline water, etc.
Solid solutions
(a) Gas in solid Solution of hydrogen in palladium
(b) Liquid in solid Amalgams, e.g., Na-Hg
(c) Solid in solid Gold ornaments (Cu/Ag with Au)

The solution likely to be formed is interstitial solid solution.

(i) Mole fraction: It is defined as the ratio of the number of moles of the solute to the total number of moles in the solution. If A is the number of moles of solute dissolved in B moles of solvent, then Mole fraction of solute

(ii) Molality: It is defined as die number of moles of a solute present in 1000g (1kg) of a solvent.

NOTE: Molality is considered better way of expressing concentration of solutions, as compared to molarity because molality does not change with change in temperature since the mass of solvent does not vary with temperature,

(iii) Molarity: It is defined as the number of moles of solute present in one litre of solution.

NOTE: Molarity is the most common way of expressing concentration of a solution in laboratory. However, it has one disadvantage. It changes with temperature because volume of a solution alters due to expansion and contraction of the liquid with temperature.

(iv) Mass percentage: It is the amount of solute in grams present in 100g of solution.

Mass of HNO3 in solution = 68 g
Molar mass of HNO3 = 63 g mol-1
Mass of solution = 100 g
Density of solution = 1·504 g mL-1

10 percent w/w solution of glucose in water means 10g glucose and 90g of water.
Molar mass of glucose = 180g mol-1 and molar mass of water = 18g mol-1

Calculation of no. of moles of components in the mixture.

15 ppm means 15 parts in million (106) by mass in the solution.

Henry’s law: The solubility of a gas in a liquid at a particular temperature is directly proportional to the pressure of the gas in equilibrium with the liquid at that temperature.
or
The partial pressure of a gas in vapour phase is proportional to the mole fraction of the gas (x) in the solution. p = KHX
where KH is Henry’s law constant.
Applications of Henry’s law :
(i) In order to increase the solubility of CO2 gas in soft drinks and soda water, the bottles are normally sealed under high pressure. Increase in pressure increases the solubility of a gas in a solvent according to Henry’s Law. If the bottle is opened by removing the stopper or seal, the pressure on the surface of the gas will suddenly decrease. This will cause a decrease in the solubility of the gas in the liquid i.e. water. As a result, it will rush out of the bottle producing a hissing noise or with a fiz.
(ii) As pointed above, oxygen to be used by deep sea divers is generally diluted with helium inorder to reduce or minimise the painfril effects during decompression.
(iii) As the partial pressure of oxygen in air is high, in lungs it combines with haemoglobin to form oxyhaemoglobin. In tissues, the partial pressure of oxygen is comparatively low. Therefore, oxyhaemoglobin releases oxygen in order to carry out cellular activities.

 Solutions having vapour pressures more than that expected from Raoult’s law are said to exhibit positive deviation. In these solutions solvent – solute interactions are weaker and ∆solH is positive because stronger A – A or B – B interactions are replaced by weaker A – B interactions. Breaking of the stronger interactions requires more energy & less energy is released on formation of weaker interactions. So overall ∆sol H is positive. Similarly ∆solV is positive i.e. the volume of solution is some what more than sum of volumes of solvent and solute.
So there is expansion in volume on solution formation.
Similarly in case of solutions exhibiting negative deviations, A – B interactions are stronger than A-A&B-B. So weaker interactions are replaced by stronger interactions so , there is release of energy i.e. ∆sol H is negative.

According to Raoult’s Law,

According to Raoult’s Law,

 Let the molar mass of solute = Mg mol-1

(i) Both w-hexane and n-octane are non-polar. Thus, the intermolecular interactions will be London dispersion forces.
(ii) Both I2 and CCl4 are non-polar. Thus, the intermolecular interactions will be London dispersion forces.
(iii) NaCl04 is an ionic compound and gives Na+ and Cl04– ions in the Solution. Water is a polar molecule. Thus, the intermolecular interactions will be ion-dipole interactions.
(iv) Both methanol and acetone are polar molecules. Thus, intermolecular interactions will be dipole-dipole interactions.
(v) Both CH3CN and C3H6O are polar molecules. Thus, intermolecular interactions will be dipole-dipole interactions.

 n-octane (C8H18) is a non-polar liquid and solubility is governed by the principle that like dissolve like. Keeping this in view, the increasing order of solubility of different solutes is:
KCl < CH3OH < CH3C=N < C6H12 (cyclohexane).

 (i) Phenol (having polar – OH group) – Partially soluble.
(ii) Toluene (non-polar) – Insoluble.
(iii) Formic acid (form hydrogen bonds with water molecules) – Highly soluble.
(iv) Ethylene glycol (form hydrogen bonds with water molecules) Highly soluble.
(v) Chloroform (non-polar)- Insoluble.
(vi) Pentanol (having polar -OH) – Partially soluble.

Mass of aspirin = 6.5 g
Mass of acetonitrile = 450 g
Then, total mass of the solution = (6.5 + 450) g = 456.5 g
Therefore, mass percentage of C9H8O4 = 456.5 × 100%
= 1.424%

Here, p = 760 mm Hg, KH = 4.27 × 105 mm Hg (at 298 K)
According to Henry’s law, p = KHχ
χ = 177.99 × 10-5
= 178 × 10-5
Hence, the mole fraction of methane in benzene is 178 × 10-5.

Fluorine being more electronegative than chlorine has the highest electron withdrawing inductive effect. Thus, triflouroacetic acid is the strongest trichloroacetic acid is second most and acetic acid is the weakest acid due to absence of any electron withdrawing group. Thus, F3CCOOH ionizes to the largest extent while CH3COOH ionizes to minimum extent in water. Greater the extent of ionization greater is the depression in freezing point. Hence, the order of depression in freezing point will be CH3COOH < Cl3CCOOH < F3CCOOH.

Mass of solute (CH3CH2CHClCOOH) = 10 g
Molar mass of
CH3CH2CHClC00H = 4 × 12 + 7 × 1 + 1 × 35.5 + 2 × 16 = 48 + 7 + 35.5 + 32
= 122.5 g mol-1
\frac{\text { Mass / Molar mass }}{\text { Mass of solvent (Kg) }} = 

 Mass / Molar mass 

 Mass of solvent (Kg) 10 g

(122.5 g mol−1)×(0.25Kg)

= 0.326 m
Let α be the degree of dissociation of CH3CH2CHClCOOH then

Total no. of moles after dissociation = 1 – α + α + α
= 1 + α
Van’t Hoff factor
Total no. of moles after dissociation

∴ i = 

1+α

1

= 1 + α
= 1 + 0.0655
= 1.0655
Hence, the depression in the freezing point of water is given as:
ΔTf = i.Kfm
= 1.0655 × 1.86 kg mol-1 × 0.326 mol kg-1
= 0.65K

Number of moles of liquid A, nA = mol = 0.714 mol
Number of moles of liquid B,nB = mol = 5.556 mol
Then, mole fraction of A, χA = (0.714+5.556)mol

 = 0.114
Mole fraction of B, χB = 1 – 0.114 = 0.886
Vapour pressure of pure liquid B, = 500 torr
Therefore, vapour pressure of liquid B in the solution,
PB = 500 × 0.886
= 443 torr
Total vapour pressure of the solution, ptotal = 475 torr
∴ Vapour pressure of liquid A in the solution,
PA = Ptotal – PB
= 475 – 443 = 32 torr
Now, PA = 280.7 torr
Hence, the vapour pressure of pure liquid A is 280.7 torr.

It can be observed from the graph that the plot for the ptotai of the solution curves downwards. Therefore, the solution shows negative deviation from the ideal behaviour.

Molar mass of benzene (C6H6) = 6 × 12 + 6 × 1 = 78g mol-1
Molar mass of toluene (C6H5CH3 ) = 7 × 12 + 8 × 1 = 92 g mol-1
No. of moles present in 80 g of benzene = 80  78

 mol = 1.026 mol
No. of moles present in 100 g of toluene = 100  92

 mol = 1.087 mol
Mole fraction of benzene, χC6H6, = 1.026+1.087

 = 0.486
∴ Mole fraction of toluene,χC6H5CH35013 = 1 – 0.486 = 0.514
It is given that vapour pressure of pure benzene, = 50.71 mm Hg
Vapour pressure of pure toluene, = 32.06 mm Hg
Therefore, partial vapour pressure of benzene,
Ptotal = χC6H6 × = 0.486 × 50.71
= 24.645 mm Hg
Partial vapour pressure of toluene, PC6H5CH3 = χC6H5CH3 ×
= 0.514 × 32.06
= 16.479 mm Hg
Total vapour pressure of solution (p) = 24.645 + 16.479
= 41.124 mm Hg
Mole fraction of benzene in vapour phase(24)mm
= 0.599 ≅ 0.6

We know that,
π = i 

RT
⇒ π = i 

RT
⇒ w = ?

Given,
π = 0.75 atm
V = 2.5L
i = 2.47
T = (27 + 273)K = 300K
R = 0.0821 L atm K-1mol–
Molar mass of CaCl2(M) = 1 × 40 + 2 × 35.5 = 111 g mol -1
Therefore, w = 

0.75×111×2.5

2.47×0.0821×300

 = 3.42g
Hence the required amount of CaCl2 is3.42g

Step I. Calculation of Van’t Hoff factor (i)
K2SO4 dissociates in water as :

 30% by mass of C6H6 in CCl4 => 30 g C6H6 in 100 g solution
.’. no. of moles of C6H6,(nC6h6) = 30/78 = 0.385

Step I. Calculation of molality of solution
Weight of KI in 100 g of the solution = 20 g
Weight of water in the solution = 100 – 20 = 80 g = 0-08 kg
Molar mass of KI = 39 + 127 = 166 g mol-1.

Step II. Calculation of molarity of solution

Step III. Calculation of mole fraction of Kl

Mass of sugar in 5% (by mass) solution means 5gin 100g of solvent (water)

1 molal solution of solute means 1 mole of solute in 1000g of the solvent.

According to Raoult’s Law,

Calculation of Van’t Hoff factor (i)
Given, w1 = 500 g = 0.5 kg, w2 = 19.5 g, Kf = 1.86 K kg mol-1, ΔTf = 1 K
Molar mass of CH2FCOOH (M2)
= 2 × 12 + 3 × 1 + 1 × 19 + 2 × 16
= 24 + 3 + 19 + 32
= 78 g mol-1
ΔTf = i Kf m
 

Percentage of oxygen (O2) in air = 20%
Percentage of nitrogen (N2) in air = 79%
Also, it is given that water is in equilibrium with air at a total pressure of 10 atm that is, (10 × 760) mm = 7600 mm
Therefore, partial pressure of oxygen,
PO2 = 1520 mm Hg
Partial pressure of nitrogen, pN2 = 6004 mm Hg
Now, according to Henry’s law,
p = KH.χ
For oxygen:


(Given KH = 6.51 × 107 mm)
= 9.22 × 10-5
Hence, the mole fractions of oxygen and nitrogen in water are 4.61 × 10-5 and 9.22 × 10-5 respectively.