Class 12 Chemistry

Solutions

NCERT Solution Part 1

Question 1: Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

Answer: Solution: A homogenous mixture of two or more than two substances is called solution. The component that is present in the largest quantity is called the solvent. The component other than the solvent is called solute.

Types of Solution:

  1. Gaseous Solution: mixture of oxygen and nitrogen gases
  2. Liquid Solution: Salt solution in water
  3. Solid Solution: Amalgam of mercury with sodium

Question 2: Give an example of a solid solution in which the solution is a gas.

Answer: Solution of hydrogen in palladium


Question 3: Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage

Mole Fraction (X): Mole fraction of a component is defined as the ratio of number of moles of the component to total number of moles of all the components.

Mole fraction of a component `=text(Number of moles of a component)/text(Total number of moles of all the components)`

Molality (m): The number of moles of the solute per kilogram of the solvent is called molality. It can be expressed as follows:

Molality (m) `=text(Moles of solute)/text(Mass of solvent in kg)`

Molarity (M): Number of moles of solute dissolved in one litre of solution is called molarity of the solute.

Molarity `=text(Moles of solute)/text(Volume of solution in litre)`

Mass Percentage (w/w): The mass percentage of a component of a solution is defined as the percentage of mass of the component in total mass of the solution.

Mass % of a component `=text(Mass of component)/text(Total mass of solution)xx100`

Question 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: Molar mass of nitric acid (HNO3 `= 1 + 14 + 3xx16=15+48=63` g mol-1

Density of solution = 1.504 g mL-1

Hence, mass of solution in 1 litre `= 1.504xx1000 = 1504` g

Mass of nitric acid `=1504xx68%=1022.72` g

Hence, moles of nitric acid `=(1022.72g) /(63g text(mol)^(-1)) =16.23` M

Molarity = 16.23 M


Question 5: A solution of glucose in water is labeled 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: Molar mass of glucose (C6H12O6) `=6xx12+12xx1+6xx16``=72+12+96=185` g mol-1

Molar mass of water = 18 g mol-1

Let us assume that total mass of solution = 100 g

Then, mass of glucose = 10 g and that of water = 90 g

In this case, number of moles of glucose `=(10)/(185)=0.054` M

And number of moles of water `=(90)/(18)=5` M

Molality can be calculated as follows:

Molality (m) `=text(Moles of solute)/text(Mass of solvent in kg)`

`m_G=(0.054M)/(0.1kg)=0.54` M kg-1

`m_w=(5M)/(0.1kg)=50` M kg-1

Mole fraction can be calculated as follows:

`X_G=(0.054M)/(0.054M+5M)=(0.054)/(5.054)=0.01`

`X_W=(5M)/(5.054M)=0.99`

Density = 1.2 g mL-1

Hence, mass of solution in 1 L = 1200 g =1.2 kg

Since 100 g contains 0.054 M of glucose and 5 M of water

So, 1200 g contains `0.054 M\xx12=0.648` M of glucose

And `5xx12=60` M of water

Molarity of glucose = 0.648 M

Molarity of water = 60 M

Question 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: Reaction between HCl and Na2CO3 can be written as follows:

2HCl + Na2CO3 ↠ 2NaCl + CO2 + H2O

This means that 2 M of HCl reacts with 1 M of Na2CO3

Reaction between HCl and NaHCO3 can be written as follows:

HCl + NaHCO3 ↠ NaCl + CO2 + H2O

This means that 1 M of HCl reacts with 1 M of NaHCO3

Molar mass of Na2CO3 `=2xx23+12+3xx16``=46+12+48=106` g mol-1

Molar mass of NaHCO3 `=23+1+12+3xx16``=36+48=84` g mol-1

Let us assume that 1 g of mixture contains x g of sodium carbonate

So, the mixture contains (`1-x`) g of sodium bicarbonate

No. of moles of sodium carbonate in x gram `=(x)/(106)`

No. of moles of sodium bicarbonate in (`1-x`) gram `=(1-x)/(106)`

As per question: `(x)/(106)=(1-x)/(84)`

Or, `84x=106-106x`

Or, `190x=106`

Or, `x=(106)/(190)=0.558` g

So, number of moles of sodium carbonate `=(0.558)/(106)=0.00526`

Number of moles of sodium bicarbonate `=(1-0.558)/(84)=0.00526`

Let us now calculate required amount of 1 M HCl

Since 1 M of sodium carbonate reacts with 2 M HCl

So, 0.00526 M of sodium carbonate will react with `0.00526xx2=0.01052` M of HCl

Since 1 M of sodium bicarbonate reacts with 1 M HCl

So, 0.00526 M of sodium bicarbonate will react with 0.00526 M of HCl

Hence, total number of moles of HCl `=0.00526+0.01052=0.01578` M

0.1 mole of 0.1 M HCl is present in 1000 mL

So, 0.01578 mole of 0.1 M HCl will be present in

`(1000)/(0.1)xx0.01578=157.8` mL


Question 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: Mass of solute in 300 g of 25% solution `=300xx25%=75` g

Mass of solute in 400 g of 40% solution `=400xx40%=160` g

Total mass after mixing `=300+400=700` g

Total mass of solute `=75+160=235` g

Mass percentage of resulting solution `=(235)/(700)xx100=33.57%`

Question 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: Molar mass of ethylene glycol C2H6O2

`=2xx12+6xx1+2xx12``=24+6+32=62` g mol-1

Number of moles of ethylene glycol `=(222.6)/(62)=3.6` M

Number of moles of water `=(200)/(18)=11.11` M

Molality of ethylene glycol `=(3.6)/(422.6)xx1000=8.52` M kg-1

Density = 1.072 mL-1

Hence, 1 litre contains 1072 g of solution

Molarity of ethylene glycol `=(3.6)/(422.6)xx1072=9.13` M L-1

Question 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).

(a) Express this in percent by mass

Answer:< 15 ppm means 15 parts in million (106) by mass in solution. Hence, this can be converted into percentage as follows:/p>

`(15)/(10^6)xx100=15xx10^(-4)%`

(b) Determine the molality of chloroform in the water sample.

Answer: Molar mass of chloroform `=12+1+3xx35``=13+105=118` g mol-1

Number of moles of chloroform `=(15)/(118)=0.13`

Molality `=(0.13)/(10^6)xx1000``=0.13xx10^3xx10^(-6)`

`=1.3xx10^(-4)` M kg-1

Question 10: What role does the molecular interaction play in a solution of alcohol and water?

Answer: Solution of alcohol and water show a large positive deviation from Raoult’s Law. This means that intermolecular attraction between alcohol and water is significantly less than that between two molecules of alcohol or water. As a result, this solution shows similar composition in liquid as well as in vapor phase. They form minimum boiling point azeotrop and it is difficult to separate them through fractional distillation.



Molality

InText Questions

NCERT Solution 1

NCERT Solution 2

NCERT Solution 3

NCERT Solution 4