Class 12 - Chemistry - Solutions

Question 2.1.

Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

Answer:

A solution is a homogeneous mixture of two or more than two substances on molecular level.

The constituent of the mixture present in a smaller amount is called the Solute & the one present in larger amount is called the Solvent.

For eg: - Small amount of sugar (solute) dissolved in water (solvent).

SOLUTE + SOLVENT = SOLUTION

There are nine types of solutions formed. They are:

State of Solute

State of Solvent

Examples

1.   Gas

Gas

Air

2.   Gas

Liquid

Oxygen in water

3.   Gas

Solid

Smoke particles in air

4.   Liquid

Gas

CO2 dissolved in water

5.   Liquid

Liquid

Alcohol in Water

6.   Liquid

Solid

Mercury in silver

7.   Solid

Gas

Adsorption of hydrogen over palladium

8.   Solid

Liquid

Sugar in water

9.   Solid

Solid

Carbon in Iron.

 

Out of these nine types of solution, solid in liquid, liquid in liquid & gas in liquid are very common.

When the components of the solution are mixed, the resulting solution may be in the solid, liquid or gaseous state.

They are:-

(i) Gaseous solution: The solution in which the solvent is a gas is called a gaseous solution.

In these solutions, the solute may be liquid, solid, or gas.

For example, a mixture of oxygen and nitrogen gas is a gaseous solution.

 

(ii) Liquid solution: The solution in which the solvent is a liquid is known as a liquid solution.

The solute in these solutions may be gas, liquid, or solid.

 

(iii) Solid solutions: The solution in which the solvent is a solid is known as a solid solution.

The solute in these solutions may be a gas, liquid or solid. For example, a solution of copper in gold is a solid solution.

 

Question 2.2.

Suppose a solid solution is formed between two substances, one whose particles are very large and the other whose particles are very small.

What kind of solid solution is this likely to be?

Answer:

The given solid solution is the example of an interstitial solid solution.

In this solution, particles of solid solute fit into voids or interstices of solvent.

For example:  tungsten carbide, where tungsten atoms are arranged in a fcc pattern with carbon atoms in octahedral holes.

 

Question 2.3.

Define the following terms:

(i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage

Answer:

(i) Mole fraction - The mole fraction of a particular component in a solution is the ratio of the

number of moles of that component to the total number of moles of all the components present in the solution.

Mathematically,

Mole Fraction of component

= (Number of moles of given component)/(Total number of moles in a solution)

Mole Fraction is independent of temperature.

(ii) Molality - Molality of a solution is defined as the number of moles of solute dissolved per 1000g [1kg] of the solvent.

It is represented by m.

Molality actually represents the concentration of solution in (mol / kg).

Mathematically,

Molality = (Number of moles of solute)/ (Mass of solvent in Kg)

It is represented by m.

(iii) Molarity- The number of moles of solute dissolved per litre of the solution at a particular temperature is called the

molarity of the solution at that temperature.

Molarity actually represents the concentration of a solution in (mol / L).

Mathematically,

Molarity = (Number of moles of solute)/ (Volume of solution in litres)

(iv) Mass percentage - Mass percentage is defined as the mass of the solute in grams dissolved per 100g of the solution.

It is also referred to as weight percentage [w/w].

For example, 10% [by mass] urea solution means that 10 g of urea are present in 100 g of solution, the solvent being only 100-10 = 90 g.

Mathematically, the mass percentage of a solute in a solution is given by:

Mass Percentage of Solute =

[(Mass of solute)/ (Mass of solute + Mass of solvent)] x100

Or

Mass Percentage of Solute = [(Mass of solute)/ (Mass of solution)] x (100)

 

Question 2.4.

Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution.

What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL–1?

Answer:

Given:

Concentration of Nitric Acid, HNO3 = 68%

Density of solution, d = 1.504 g/ml

Molarity = Mo = (Number of moles of solute)/ (Volume of solution in litres)

Density, d = (Mass [M])/ (Volume [V])

68% of Nitric acid by mass in aqueous solution means that 68g [[68 × 100]/100] of Nitric acid present in 100g of solution.

⇒ Molecular mass of Nitric Acid, HNO3 = [1 × 1] + [1 × 14] + [16 × 3]

= 63g

⇒Number of moles of Nitric Acid = [68/63]

= 1.079 moles

⇒ Given Density, d = 1.504 g/ml

⇒ Volume, v = [100/1.504]

= 66.489 ml

⇒ Molarity, Mo = [1.079/66.489] × 1000

= 16.23 M

Therefore the molarity of the sample is 16.24 M

 

Question 2.5.

A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution?

If the density of solution is 1.2 g mL–1, then what shall be the molarity of the solution?

Answer:

10% w/w solution of glucose in water means that 10 g of glucose in present in 100 g of the solution i.e.,

10 g of glucose is present in (100 - 10) g = 90 g of water.

Molar mass of glucose (C6H12O6) = (6 × 12 + 12 × 1 + 6 × 16) = 180 g mol-1

Then, number of moles of glucose = (10 / 180) mol

= 0.056 mol

Therefore, Molality of solution = (0.056 mol / 0.09kg) = 0.62 m

Number of moles of water = (90g / 18g mol-1) = 5 mol

Mole fraction of glucose (xg) = (0.056) / (0.056+5) = 0.011

And, mole fraction of water xw = (1 – xg)

= (1 - 0.011) = 0.989

If the density of the solution is 1.2 g mL-1, then the volume of the 100 g solution can be given as:

= (100g / 1.2) g mL-1

= 83.33 mL

=83.33 x 10-3 L

Therefore, Molarity of the solution = (0.056 mol / 83.33 x 10-3) L

= 0.67 M

 

Question 2.6.

How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both?

Answer:

Molar mass of Na2CO3 = 106

Molar mass of NaHCO3 = 84

 1 moles of each makes (106 + 84) g = 190 g

 For a equimolar mixture:

190 g of mixture contains 1 moles of each

1   g of mixture contains (1/190) moles of each = 5.26 X 10-3 mole of each

 Now neutralisation reaction is as below:

 Na2CO3 +   2HCl --> 2 NaCl + CO2 + H2O

   1 mol   2 mol

  (1/190) mol (2/190) mol

 NaHCO3   +   HCl  --> NaCl + CO2 + H2O

1 mol        1 mol

 (1/190) mol (1/190) mol

 So to neutralise 1 g of mixture HCl required

= ((2/190) + (1/190)) mol = (3/190) mol

 1 mol HCl is present in 1 M solution having 1000 ml volume

1 mol HCl is present in 0.1 M solution having (1000/0.1) ml volume

(3/190) mol HCl is present in 0.1 M solution having (1000/0.1) x (3/190) ml volume = 157.89 ml

 So 157.89 ml of 0.1M HCl is required.

 

Question 2.7.

A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass.

Calculate the mass percentage of the resulting solution.

Answer:

Total amount of solute present in the mixture is given by= (300 x (25/100) + 400)

= (75 + 160)

=235 grams

Total amount of solution = (300 + 400) = 700 gms

Therefore, mass percentage (w/w) of the solute in the resulting solution

= [(235/700)] x 100% = 33.57%

And, mass percentage (w/w) of the solvent in the resulting solution

= (100 – 33.57) %

=66.43%

                                                                                                                                                                                   

Question 2.8.

An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water.

Calculate the molality of the solution. If the density of the solution is 1.072 g mL–1, then what shall be the molarity of the solution?

Answer:

Calculation of Molality:

Mass of ethylene glycol = 222.6 (Given)

Molar mass of ethylene glycol [C2H4 (OH) 2]

= (2 X 12 + 6 x 1 + 2 x 16)

= 62

Therefore moles of ethylene glycol

= (222.6g / 62) gmol-1

= 3.59 mol

Mass of water = 200g   (Given)

Therefore molality of the solution is

= (moles of ethylene glycol / mass of water) x (1000)

= (3.59 / 200) x 1000

= 17.95 m

Calculation of Molarity:

Moles of ethylene glycol = 3.59 mol

Total Mass of solution = (200 + 222.6)

= 422.6g

Volume of solution = (mass / density volume)

= (422.6 / 1.072)

= 394.22 ml

So the molarity of the solution  

= (moles of ethylene glycol / volume of solution) x (1000)

= (3.59 / 394.22) x (1000)

= 9.11 M

 

Question 2.9.

A sample of drinking water was found to be severely contaminated with chloroform (CHCl3) supposed to be a carcinogen.

The level of contamination was 15 ppm (by mass):

(i) Express this in percent by mass (ii) determine the molality of chloroform in the water sample.

Answer:

15 ppm means 15 parts of Chloroform in 106 parts of drinking water

Therefore, Mass Percentage = [(Mass of chloroform)/ (total mass)] x (100)

= (15/106) x (100)

= 1.5 × 10-3

Calculation of Molality:

Molecular Mass of Chloroform, CHCl3 = [12] + [1] + [35.5 × 3]

=119.5 g

Number of Moles of Chloroform = [15 / 119.5]

= 0.1255 moles

Molality = (Number of moles)/Mas of solvent in kg)

= (0.1255/106) x (1000)

= 1.255 × 10-4

Therefore the Mass Percentage is = 1.5 × 10-3 and the Molality of the solution

= 1.255 × 10-4 m.

 

Question 2.10.

What role does the molecular interaction play in a solution of alcohol and water?

Answer:

The lower members of alcohols are highly soluble in water but the solubility decreases with increase in the molecular weight.

The solubility of lower alcohols in water is due to formation of hydrogen bonds (Hydrogen bonding) between alcohols & water molecules.

Hydrogen bonding

However, as the size of alcohol molecule increases, the alkyl groups becomes larger & prevents the formation of

hydrogen bonds with water, & hence the solubility goes on decreasing with increase in the length of carbon chain.

Also the interaction between the molecules of alcohol and water is weaker than alcohol−alcohol and water−water interactions.

As a result, when alcohol and water are mixed, the intermolecular interactions become weaker and the molecules can easily escape.

This increases the vapour pressure of the solution, which in turn lowers the boiling point of the resulting solution.

 

Question 2.11.

Why do gases always tend to be less soluble in liquids as the temperature is raised?

Answer:

Whenever a gas is dissolved in a liquid, a small amount heat is liberated in the process.

So dissolving a gas in liquid is overall an exothermic process.

So according to the LeChatelier principle, whenever the temperature is increased for a reaction which is exothermic in nature,

the equilibrium shifts backwards and the reaction proceeds in backward direction that means the solution gets dissociated and will give off gas

and hence solubility of gas decreases.

So with the increase in temperature, the solubility of the gases in liquids decreases.

 

 

Question 2.12.

State Henry’s law and mention some important applications?

Answer:

Henry’s law states that the mass of a gas dissolved per unit volume of the solvent at a given temperature is proportional

to the pressure of the gas in equilibrium with the solution.

Or

It also states that the pressure of a gas over a solution in which the gas is dissolved is proportional to the

mole fraction of the gas dissolved in the solution.

The important applications of Henry’s law are as follows:

1) In the production of carbonated beverages-in order to increase the solubility of CO2 in cold drinks, beer etc.,

they are sealed under high pressure. When the bottle is opened under normal atmospheric pressure,

the pressure inside the bottle falls to atmospheric pressure & the excess CO2 bubbles out of the bottle causing effervescence.

 

2) At high altitudes-the partial pressure of oxygen at high altitudes is less than the ground level.

This results in low concentration of oxygen in the blood & tissues of the peoples

3) In scuba diving- during scuba diving, when the diver breaths in compressed air from the supply tank,

more nitrogen dissolves in the blood & other body fluids because the pressure at that depth is

far greater than the surface atmospheric pressure.

 

 

Question 2.13.

The partial pressure of ethane over a solution containing 6.56 × 10–3 g of ethane is 1 bar.

If the solution contains 5.00 × 10–2 g of ethane, then what shall be the partial pressure of the gas?

Answer:

Molar mass of ethane (C2H6) = (2 × 12 + 6 × 1)

= 30 g mol−1

Number of moles present in 6.56 × 10−2 g of ethane

= 2.187 × 10−4 mol

Let the number of moles of the solvent be x.

According to Henry’s law,

p = KH x

1 bar = KH (2.187 x 10-4)/ (2.187 x 10-4 +x)

Or 1 bar = KH (2.187 x 10-4)/(x) (because x>> 2.187 x 10-4)

  • KH = (x) /(2.187 x 10-4) bar

Number of moles present in 5.00 × 10−2 g of ethane

= (5.00 x 10-2) / (30) mol

= 1.67 × 10−3 mol

According to Henry’s law,

p = KH x

=(x) / (2.187 x 10-4) x (1.67 x 10-3)/ [(1.67 x 10-3) + (x)]

= [(x)/ (2.187 x 10-4) x (1.67 x 10-3)/(x)] (Because x >> 1.67 x 10-3)

= 7.636 bar

Hence, partial pressure of the gas shall be 7.636 bars.

 

Question 2.14.

What is meant by positive and negative deviations from Raoult's law and how is the sign of ΔmixH related to positive

and negative deviations from Raoult's law?

Answer:

Raoult’s law states that at a given temperature, the vapour pressure of a solution containing non-volatile solute

is directly proportional the mole fraction of the solvent.

Non-ideal solutions show positive & negative deviations from ideal behaviour.

Non-ideal solutions showing  positive deviations from Raoult’s law- Consider a binary solution of two components A & B .

If the A-B interaction in the solutions are weaker than A-A & B-B interactions in the two liquids forming the solution,

then the escaping tendency of A & B  types of molecules from the solution becomes more than from pure liquids.

As a result, each component of solution has a partial vapour pressure greater than expected on the basis of Raoult’s law.

This is called positive deviations from Raoult’s law, i.e. PA> PA °xA & PB >PB°xB

 

 

Question 2.15.

An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bars at the normal boiling point of the solvent.

What is the molar mass of the solute?

 

Answer:

Vapour pressure of the solution at normal boiling point (p1) = 1.004 bar (Given)

Vapour pressure of pure water at normal boiling point (p10) = 1.013 bar

Mass of solute, (w2) = 2 g

Mass of solvent (water), (w1) = (100 – 2) = 98 g

Molar mass of solvent (water), (M1) = 18 g mol-1

According to Raoult's law,

(p10 - p1) / (p10)   = (w2 x M1) / (M2 x w1)

(1.013 - 1.004) / (1.013) = (2 x 18) / (M2 x 98)

(0.009 / 1.013)  =   (2 x 18) / (M2 x 98)

M2 = (2 x 18 x 1.013) / (0.009 x 98)

M2 = 41.35 g mol- 1

Hence, the molar mass of the solute is 41.35 g mol - 1.

 

 

Question 2.16.

Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively.

What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?

Answer:

Vapour pressure of heptane p10 = 105.2 kPa

Vapour pressure of octane p20= 46.8 kPa

Also, Molar mass of heptane (C7H16) = (7 × 12 + 16 × 1) = 100 g mol-1

Therefore, Number of moles of heptane = (26/100) mol = 0.26 mol

Molar mass of octane (C8H18) = (8 × 12 + 18 × 1) = 114 g mol-1

Therefore, Number of moles of octane = (35/114) mol = 0.31 mol

Mole fraction of heptane, x1 = (0.26) / (0.26 +0.31)

= 0.456

And, mole fraction of octane, x2 = (1 - 0.456) = 0.544

Now, partial pressure of heptane, p1 = x2 p20

= (0.456 × 105.2)

= 47.97 kPa

Partial pressure of octane, p2 = x2 p20

= (0.544 × 46.8) = 25.46 kPa

Hence, vapour pressure of solution, ptotal = (p1 + p2)

= (47.97 + 25.46)

= 73.43 kPa

 

Question 2.17.

The vapour pressure of water is 12.3 kPa at 300 K.

Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Answer:

1 molal solution means 1 mole of solute present in 1000g of water solvent)

Molecular weight of water = H2O = (1 × 2 + 16) = 18g/mol

No. of moles of water, n = (given mass) / (molecular weight)

=> n = (1000/18) = 55.56 gmol-1

Mole fraction of solute in solution, x2 = (moles of solute)/ [(moles of solute + moles of water)]

Or x2 = (1)/ (1 + 55.56)

⇒ x2 = 0.0177

Given vapour pressure of pure water at 300k is 12.3 kPa

From formula:-

(p10 - p1) / (p10)   = x2

(12.3 - p1)/ (12.3) = 0.0177

Or p1 = 12.0823 kPa

Itis the vapour pressure of the solution.

 

Question 2.18.

Calculate the mass of a non-volatile solute (molar mass 40 g mol–1) which should be dissolved in 114 g

octane to reduce its vapour pressure to 80%.

Answer:

Let the vapour pressure of pure octane be p10.

Then, the vapour pressure of the octane after dissolving the non-volatile solute is (80/100) p10 = 0.8 p10.

Molar mass of solute, M2 = 40 g mol-1

Mass of octane, w1 = 114 g

Molar mass of octane, (C8H18), M1 = (8 × 12 + 18 × 1) = 114 g mol-1

Applying the relation,

(p10 - p1) / (p10)    = (w2 x M1) / (M2 x w1)

(p10 - 0.8 p10) / (p10)    = (w2 x 114) / (40 x 114)

0.2 (p10 / p10)   = (w2 / 40)

⇒ .2 = (w2 / 40)

w2 = 8 g

Hence, the required mass of the solute is 8 g.

 

 

Question 2.19.

A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K.

Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:

(i) molar mass of the solute (ii) vapour pressure of water at 298 K.

Answer:

Let, the molar mass of the solute be Mg mol- 1

Now, the no. of moles of solvent (water), n1 = (90g / 18g) mol-1

And, the no. of moles of solute, n2 = (30g / M mol-1) = (30 / M mol)

p1 = 2.8 kPa

Applying the relation:

(p10 - p1) / (p10)    = (n2) / (n1 + n2)

 (p10 - 2.8) / (p10)    = (30/M) / {5 + (30/M)}

=> (1) - (2.8/p10) = (30/M) / {(5M+30)/M}

=> (1) - (2.8/p10) = (30) / (5M + 30)

=> (2.8/p10) = (1) - (30) / (5M + 30)

=> (2.8/p10)   = (5M + 30 - 30) / (5M + 30)

=> (2.8)/ (p10)   = (5M) / (5M+30)

=> (p10) / (2.8) = (5M+30) / (5M) ---------------- (1)

After adding 18 g of water:

n1 = (90+18g) / (18) = 6 mol

And the new vapour pressure is p1 = 2.9 kPa   (Given)

Again, using the relation:

(p10 - p1) / (p10)    = (n2) / (n1 + n2)

=> (p10 - 2.9) / (p10)    = (30/M) / {6 + (30/M)}

=> (1) - (2.9/p10) = (30/M) / {(6M+30)/M}

=> (1) - (2.9/p10) = (30) / (6M + 30)

=> (2.9/p10) = (1) - [(30) / (6M + 30)]

=> (2.9/p10)   = (6M + 30 - 30) / (6M + 30)

=> (2.9/p10)   = (6M) / (6M+30)

=> (p10 / 2.9) = (6M+30) / (6M) ---------------- (2)

Dividing equation (1) by (2), we get:

(2.9 / 2.8) =   {(5M+30) / 5M} / {(6M+30) / 6M}

=> (2.9) x (6M+30 / 6) = (5M+30 / 5) x 2.8

=> 2.9 x (6M +30) x 5 = (5M+30) x 2.8 x 6

=> (87M + 435) = (84M + 504)

=> 3M = 69

=> M = 23u

Therefore, the molar mass of the solute is 23 g mol- 1.

(ii) Putting the value of 'M' in equation (i), we get:

=> (p10) / (2.8) = (5M+30) / (5M)

=> (p10) / (2.8) = (5x23+30) / (5x23)

=> p10 = (145 x 2.8) / (115)

=> p10 = 3.53

Hence, the vapour pressure of water at 298 K is 3.53 kPa.

 

Question 2.20.

A 5% solution (by mass) of cane sugar in water has freezing point of 271K.

Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.?

Answer:

Given:

Freezing point of solution =271K

Let total mass of solution = 100gram

Then 5% of 100 gram of solution contains 5 g of solute

Now,

Change in freezing point

∆Tf = (freezing point of water – freezing point of solution)

= (273.15 – 271)

= 2.15K

Molar mass of glucose (C6H12O6) = 180 g/mol

Molar mass of cane sugar (C12H22O11) =342g/mol

Number of moles = (given mass/molar mass)

Number of moles of (C6H12O6) = (5/180) = 0.028 moles

Number of moles of (C12H22O11) = (5/ 342) = 0.0146 moles

Mass of solvent = (total mass – mass of solute)

 = (100 – 5) = 95 g or 0.095Kg

Molality = (number of moles of solute/ mass of solvent in kg)

Molality of cane sugar = (0.0146/0.095) = 0.154m

Molality of C6H12O6 (m) = (0.028/0.095) =0.29m

Deviation from freezing point is given by

∆Tf = (Kf × molality)

Putting values for cane sugar:

2.15= (Kf ×0.154)

Kf = 13.97

Putting value of glucose

∆Tf = (Kf × molality)

∆Tf = (13.97× 0.29)

∆Tf = 4.08K

Now freezing point of 5% glucose in water = (freezing point of water - ∆Tf) = (273.15 - 4.08)

= 269.07K

 

Question 2.21.

Two elements A and B form compounds having formula AB2 and AB4.

When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K.

The molar depression constant for benzene is 5.1 K kg mol–1.

Calculate atomic masses of A and B.?

Answer:

MB = (Kf x wB x 1000) / (wA X ΔTf)

Now ΔTf = 2.3, wB = 1.0, wA = 20, KF = 5.1 (given)

Putting the values in the equation

MB = (5.1 x 1 x 1000) / (20 x 2.3) = 110.87 g/mol

Therefore MAB2 = 110.9

For AB4 compound

ΔTf = 1.3, wB = 1, wA = 20

MB = (5.1 X 1 X 1000) / (20 X 1.3) = 196 g/mol

Therefore MAB = 196

Let x be the atomic mass of A & y be the atomic mass of B,

Therefore, MAB2 = (x + 2y) = 110.9 ---------------------------------- (1)

And MAB = x + 4y = 196        ---------------------------------- (2)

Subtracting equation 2 from 1, we get

2y = (196-110.9)

y   = (85.1 / 2)

y = 42.6

Putting the value of y in 1 we get

x = (110.9 – 2) x (42.6)

x = 25.59

Therefore atomic mass of A = 25.59 u Atomic mass of B = 42.6 u.

 

Question 2.22.

At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bars.

If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

Answer:

Given,

T = 300 K

π = 1.52 bar

R = 0.083 bar L

Applying the relation, π = CRT

Where

π = osmotic pressure of solution

C = concentration of solution

R = universal gas constant

T = temperature

  • C= (π /RT)
  • C = 0.061mol/L

Concentration of the solution is 0.061mol/L

 

 

Question 2.23.

Suggest the most important type of intermolecular attractive interaction in the following pairs.

(i) n-hexane and n-octane

(ii) I2 and CCl4

(iii) NaClO4 and water

(iv) Methanol and acetone

(v) Acetonitrile (CH3CN) and acetone (C3H6O).

Answer:

(i) Both the compounds are non-polar and they do not attract each other because they do not form any polar ions.

Vanderwaal’s forces of attraction will be dominant in between them as vanderwaal’s forces of attraction are not a result of any chemical or electronic bond.

(ii) Here both the compounds are non-polar because in I2 both the atoms are same so they have same electronegativity and

hence there will be no displacement of electron cloud, it will be in the centre. In case of CCl4 molecule,

it has tetrahedral shape so two Cl atoms will cancel the attraction effect from two opposite Cl atoms, hence molecule as a whole is non-polar.

Therefore they will also have vanderwaal’s forces of attraction.

(iii) NaClO4 is ionic in nature as Na, Cl and O all have different electronegativity and their shape is also not symmetric.

So molecular will be ionic in nature and we know that water is polar because O will attract the electron cloud towards it (more electronegative)

hence there will be formation of dipole (two oppositely charged ions separated by a short distance).

Therefore there will be ion-dipole interaction between them.

(iv) Methanol and acetone are both polar molecules because of the presence of electron withdrawing O atom in methanol and ketone group in acetone.

So they will have dipole-dipole interaction.

(v) Acetonitrile is polar compound due to presence of electronegative N atom and acetone due to ketone group.

So, dipole-dipole interaction.

  

Question 2.24.

Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain.

Cyclohexane, KCl, CH3OH, CH3CN

Answer:

Here n-octane is non-polar solvent due to long chain saturated structure.

We know that “like dissolves like” so a non-polar compound will be more soluble in non-polar solvent as compared to polar compound.

So cyclohexane is non-polar due to symmetric structure. KCl is ionic in nature as it will dissociate into K + and Cl- ions.

CH3CN is polar as mentioned above and CH3OH is also polar in nature.

The order of increasing polarity is:

Cyclohexane < CH3CN < CH3OH < KCl (O is more electronegative than N)

Therefore, the order of increasing solubility is:

KCl < CH3OH < CH3CN < Cyclohexane

 

Question 2.25.

Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?

  • phenol
  • toluene
  • formic acid
  • ethylene glycol
  • chloroform

(vi)  Pentanol.

Answer:

Water is a polar compound (due to electronegativity difference between O and H). We know that “like dissolves like”.

So, a non-polar compound will be more soluble in non-polar solvent as compared to polar compound.

(i) Phenol has the polar group -OH and non-polar group –C6H5 and it cannot form H bonding with water (presence of bulky non-polar group).

Thus, phenol is partially soluble in water.

(ii) Toluene has no polar groups. Thus, toluene is insoluble in water.

(iii) Formic acid (HCOOH) has the polar group -OH and can form H-bond with water.

Thus, formic acid is highly soluble in water.

(iv) Ethylene glycol(OH-CH2-CH2-OH) has polar -OH group and can form H-bond with water.

Thus, it is highly soluble in water.

(v) Chloroform is partly soluble as CHCl3 is polar in nature due to high electronegativity of Cl atoms, there will be production of partial + charge

on H atom so it can form H bonding with water but it is also surrounded by 3 Cl atoms, so partly soluble.

(vi) Pentanol (C5H11OH) has polar -OH group, but it also contains a very bulky non-polar group –C5H11.

Thus, pentanol is partially soluble in water.

 

 

Question 2.26.

If the density of some lake water is 1.25g mL–1 and contains 92 g of Na+ ions per kg of water, calculate the molality of Na+ ions in the lake.

Answer:

Mass of ions = 92g

Molar mass of ions = Na+ = 23g (neglect the mass lost due to absence of an electron)

Moles of ions = (mass of ions)/ (molar mass)

n = (92/23) moles

n = 4moles

Molality of solution = (moles of solute)/ (mass of solvent (in kg))

Molality = (4/1) = 4M

 

Question 2.27.

If the solubility product of CuS is 6 × 10–16, calculate the maximum molarity of CuS in aqueous solution.

Answer:

The Solubility product of CuS (ksp) = 6 × 10-16

CuS → Cu 2+ + S2-

Let the s be solubility of CuS in mol/L

Ksp = [Cu 2+] [S2]

Ksp = solubility product

(6 × 10-16) = s × s = s2

s=√ (6 × 10-16)

S = (2.45 × 10-8) mol/L

Hence, the maximum molarity of CuS in an aqueous solution is 2.45 × 10-8 mol/L

 

Question 2.28.

Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.

Answer:

Mass of aspirin (C9H8O4) = 6.5g (given)

Mass of acetonitrile (CH3CN) = 450g (given)

Now total mass of the solution = (6.5 + 450) = 456.5g

Therefore mass percentage of aspirin (C9H8O4)

(6.5 / 456.5) x (100)

= 1.42 %

 

 

Question 2.29.

Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in narcotic users.

Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 – 10–3 m aqueous solution required for the above dose.

Answer:

Molecular mass of nalorphene (C19H21NO3),

= (19 x 12 + 21 x 1 + 1 x 14 + 3 x 16) = 311 g/mol

Moles of nalorphene (C19H21NO3) = (1.5 x 10-3 x 311) = 4.82 x 10-6 mol

Now molality = (moles of solute / mass of solvent in g) x (1000)

Putting the values in above equation, we get

1.5 x 10-3 = (4.82 x 10-6 / mass of water) x (1000)

Or

Mass of water = (4.82 x 10-6 x 1000) / (1.5 x 10-3)

Therefore mass of water = 3.2g

 

Question 2.30.

Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.

Answer:

Molarity = (moles of the solute / volume of solution)

Putting the given values in above equation, we get

0.15 = (mole of benzoic acid / 250) x (1000)

Or

Moles of benzoic acid = (0.15 x 250) / (1000)

= 0.0375 mol of benzoic acid

Also molecular mass of benzoic acid (C6H5COOH)

= (7 × 12 + 6 × 1 + 2 × 16)

 

= 122 g/mol

Therefore amount of benzoic acid = (0.0375 x 122) = 4.575 g

 

 

Question 2.31.

The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases

in the order given above. Explain briefly.

Answer:

The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and

trifluoroacetic acid increases in the pattern,

Acetic acid< trichloroacetic acid< trifluoroacetic acid

This is because fluorine is more electronegative than chlorine.

So, trifluoroacetic acid is a stronger acid in comparison to trichloroacetic acid and acetic acid.

And also, acetic acid is the weakest of all.

Explanation: Stronger acid produces more number of ions; therefore it has more ΔTf (depression in freezing point), hence lower freezing point.

As the acidic strength increases, the acid gets more and more ionised.

Trifluoroacetic acid ionizes to the largest extent.

Hence, in this case, trifluoroacetic acid being the strongest acid produces more number of ions (extent of ionisation and concentration of ions are more),

high ΔTf (depression in freezing point) and lower freezing point and vice versa.

 

 

Question 2.32.

Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 × 10–3, Kf = 1.86 K mol–1

Answer:

Molar mass of CH3CH2CHClCOOH

= (15 + 14 + 13 + 35.5 + 12 + 16 + 16 + 1)

= 122.5 g/mol

Therefore Moles of CH3CH2CHClCOOH = (10g / 122.5) g/mol

= 0.0816 mol

Therefore molality of the solution

= (0.0816 x 1000) / (250)

= 0.3265 mol kg-1

Now if a is the degree of dissociation of CH3CH2CHClCOOH,

                                 CH3CH2CHClCOOH CH3CH2CHClCOO- + H+

Initial Concentration C molL-1                     0                           0

At equilibrium             C (1- α)                C α                          C α

So, Ka = (Cα x Cα) / (C (1-α))

Ka = (Cα2)/ (1-α)

Since α is very small with respect to 1, (1 – α) = 1

Ka = Cα2

α = √Kα / (C)

Putting the values, we get

α =   √ (1.4 x 10-3) / (0.3265)

= 0.0655

Now at equilibrium, the van’t hoff factor i = (1-α +α + (α/1))

= (1 + 0.0655)

= 1.0655

Hence, the depression in the freezing point of water is given as:

Therefore ΔTf = (i Kf m v)

= (1.065 v x 1.86 x 0.3265)

= 0.647°

 

Question 2.33.

19.5 g of CH2FCOOH is dissolved in 500 g of water.

The depression in the freezing point of water observed is 1.00 C.

Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.

Answer:

Molecular mass of CH2FCOOH

= (14 + 19 + 12 + 16 + 16 + 1) = 78 g/mol

Now, Moles of CH2FCOOH = (19.5 / 78)

= 0.25

Taking the volume of the solution as 500 mL, we have the concentration:

C = (0.25 / 500) X (1000)

Therefore Molality = 0.50m

So now putting the value:

 

ΔTf = (Kf x m)

= (1.86 x 0.50) = 0.93K

Van’t hoff factor

= (observed freezing point depression / calculated freezing point depression)

= (1 / 0.93) = 1.0753

Let α be the degree of dissociation of CH2FCOOH

                                 CH2FCOOH CH2FCOOH - + H+

Initial Concentration C molL-1                     0            0

At equilibrium             C (1- α)                C α            C α

 

Now total number of moles = (m (1-a) + ma +ma) = m (1+a)

Or

i = (α) (1+α) / (α)

= (1 +α) = 1.0753

Therefore α = (1.0753- 1)

= 0.0753

Now the Value of Ka is given as:

Ka = [CH2FCOO-] [H+] / (CH2FCOOH)

= (Cα x Cα) / (C (1-α))

= (Cα2) / (1-α)

Ka = (0.5 X (0.0753)2) / (1-0.0753)

= (0.5 X 0.00567) / (0.09247)

= 0.00307 (approx.)

= 3 X 10-3

 

Question 2.34.

Vapour pressure of water at 293 K is 17.535 mm Hg.

Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Answer:

Vapour pressure of water, p1° = 17.535 mm of Hg

Mass of glucose, w2 = 25 g

Mass of water, w1 = 450 g

Also,

Molar mass of glucose (C6H12O6),

M2 = (6 × 12 + 12 × 1 + 6 × 16) = 180 g mol-1

Molar mass of water, M1 = 18 g mol-1

Then, number of moles of glucose, n1 = (25/180) = 0.139 mol

And, number of moles of water, n2 = (450/18) = 25 mol

Also, we know that,

(p1° - p°) / (p1°) = (n1) / (n2 + n1)

=> (17.535 - p°) / (17.535) =   (0.139) / (0.139+25)

=> (17.535 - p1) = 0.097

=> p1 = 17.44 mm of Hg

Hence, the vapour pressure of water is 17.44 mm of Hg

  

Question 2.35.

Henry’s law constant for the molality of methane in benzene at 298 K is

4.27 × 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Answer:

Henry’s law constant KH = 4.27X 105 mm Hg,

p = 760mm Hg,

Using Henry’s law,

p= (KH) x (xgas)

xgas = (KH/p)

760 = (4.27x 105) x (x)

Or

xgas = (760) / (4.27x 105)

xgas = 177.99 x 10-5

 

Question 2.36.

100 g of liquid A (molar mass 140 g mol–1) was dissolved in 1000 g of liquid B (molar mass 180 g mol–1).

The vapour pressure of pure liquid B was found to be 500 Torr.

Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total

vapour pressure of the solution is 475 Torr.

Answer:

Number of moles of liquid A, nA = (100/140) mol

=0.714 mol

Number of moles of liquid B, nB = (100/180) mol

=5.556 mol

Then mole fraction of A, xA = (nA)/ (nA + nB)

= (0.714)/ (0.714 + 5.556)

=0.114

And mole fraction of B = (xB -1) = (1 -0.114)

=0.886

Also,

ptotal = (pA + pB)

= (poA xA+ poB xB)

475 = (poA x 0.114) + (500 x0.886)

poA = 280.7 Torr

Hence, vapour pressure of pure A = 280.7 Torr

Vapour pressure of A in solution = (280.7 x 0.114)

=32 Torr

Now,

pA = poA xA

poA = (pA /xA)

= (32/0.114)

=280.7 Torr

 

 Question 2.37.

Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively.

Assuming that they form ideal solution over the entire range of composition, plot ptotal, pchloroform, and pacetone as a function of xacetone.

The experimental data observed for different compositions of mixture is:

100 x xacetone

0

11.8

23.4

36.0

50.8

58.2

64.5

72.1

p acetone /mm Hg     

0

54.9

110.1

202.4

322.7

405.9

454.1

521.1

 

p chloroform /mm Hg  

632.8

548.1

469.4

359.7

257.7

193.6

161.2

120.7

 

          

Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.

Answer:

100 x xacetone

0

11.8

23.4

36.0

50.8

58.2

64.5

72.1

p acetone /mm Hg     

0

54.9

110.1

202.4

322.7

405.9

454.1

521.1

 

p chloroform /mm Hg  

632.8

548.1

469.4

359.7

257.7

193.6

161.2

120.7

 

 

Class_12_Solutions_Graph1

It can be observed from the graph that the plot for the p total of the solution curves downwards.

Therefore, the solution shows negative deviation from the ideal behaviour.

 

Question 2.38.

Benzene and toluene form ideal solution over the entire range of composition.

The vapour pressure of pure benzene and naphthalene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively.

Calculate the mole fraction of benzene in vapour phase if 80 g

of benzene is mixed with 100 g of naphthalene.

Answer:

Molar mass of benzene (C6H6) = (6 X 12) + (6 X 1) = 78 g/mol

Molar mass of toluene = (7 x 12) + (8 x 1) = 92 g/mol

Now no of moles in 80g of benzene = (80 / 78) = 1.026 mol

No of moles in 100g of toluene = (100 / 92) = 1.087 mol

Therefore, Mole fraction of benzene xb = (1.026 / 1.026) + (1.087) = 0.486

And Mole fraction of toluene, xt = (1 - 0.486) = 0.514

Also given that

Vapor pressure of pure benzene pb° = 50.71 mm Hg

And, vapour pressure of pure toluene, pt° = 32.06 mm Hg

Therefore partial Vapor pressure of benzene, pb = (pb X xb)

= (50.71 x 0.486)

= 24.65 mm Hg

And partial Vapor pressure of toluene, pt = (pt X xt)

Pt = (pt° X xt) = (32.06 x 0.514)

= 16.48

Total vapour pressure = (24.65 + 16.48) = 41.13 mm Hg

Mole fraction of benzene in vapour phase = (24.65 / 41.13) = 0.60

 

 

Question 2.39.

The air is a mixture of a number of gases.

The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K.

The water is in equilibrium with air at a pressure of 10atm.

At 298 K if the Henry’s law constants for oxygen and nitrogen at 298 K are 3.30 × 107 mm and

6.51 × 107 mm respectively, calculate the composition of these gases in water.

Answer:

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 Hg = 7600 mm Hg

Therefore, Partial pressure of oxygen, po2 = (20/100) x (7600)

=1520 mm Hg

Partial pressure of nitrogen, pN2 = (79/100) x (7600)

= 6004 mmHg

Now, according to Henry's law:

p = KH x

For oxygen:

po2 = KH xO2

=>xO2 = (po2 / KH)

= (1520 / 3.30) X (107)

= (4.61x 10-5)

For nitrogen:

pN2 = KH xN2

or xN2 = (pN2 / KH)

= (6004 / 6.51) x (107)

= 9.22 x 10-5

Hence, the mole fractions of oxygen and nitrogen in water are 4.61 ×10-5 and 9.22 × 10-5 respectively.

 

Question 2.40.

Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27° C.

Answer:

We know that

π = (i (n/V) RT)

=>π = i (w/MV) iRT

=> w = (πMV) / (iRT)     ....................... (1)

Now we have given below values:

π = 0.75 atm

V = 2.5L

i = 2.47

T = (27+273) K = 300K

Here,

R = 0.0821L atm k-1 mol-1

M = (1x40 + 2x35.5)

= 111 g/mol

Now putting the value in equation 1:

w = (0.75x111x2.5) / (2.47x0.0821x300)

=3.42g

Hence, the required amount of CaCl2 is 3.42 g.

 

Question 2.41.

Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25° C,

assuming that it is completely dissociated.

Answer:

Given that

Mass of K2SO4, w = 25 mg = 0.025g (using 1 g = 1000 mg)

Volume V = 2 liter

T = (25 + 273) = 298 K (add 273 to convert in Kelvin)

The reaction of dissociation of K2SO4

K2SO4 → 2K+ + SO42-

 Number if ions produced = (2+ 1) = 3

So van’t Hoff factor i = 3

Use the formula of Osmotic pressure

π = i (n/v) RT

=i (w/M) (1/v) RT

Gas constant, R = 0.0821 L atm K-1mol-1

Molar mass of K2SO4, M = (2 × 39 + 1 × 32 + 4 × 16) = 174 g mol-1

Substituting the values we get

π = (3) x (0.025/174) x (1/2) x (0.0821) x (298)

π= 5.27 × 10-3 atm

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