The document Solved Examples - Solutions, Class 12, Chemistry | EduRev Notes is a part of the Class 12 Course Chemistry Class 12.

All you need of Class 12 at this link: Class 12

**Example ****1. The molarity of 20% (W/W) solution of sulphuric acid is 2.55 M. The density of the solution is: ****(a) 1.25 g cm ^{-3}**

Volume of 100 g of solution = 100/d ml

= 1.249 â‰ˆ 1.25

H_{2}SO_{4}

Volume of 100 gram of the solution = 100/d

Number of moles of H_{2}SO_{4} in 100 gram of the solution = 13/98**= 1.445 M****Example ****3. Calculate the molarity of pure water (d = 1g/L)****(a) 555 M****(b) 5.55 M****(c) 55.5 M****(d) None****Ans. (c)****Solution.**

Consider 1000 mL of water

Mass of 1000 mL of water

= 1000 Ã— 1 = 1000 gram

Number of moles of water = 100/18

= 55.5

= 55.5/1**= 55.5 M****Example ****4. Calculate the quantity of sodium carbonate (anhydrous) required to prepare 250 ml of 0.1 M solution-****(a) 2.65 gram****(b) 4.95 gram****(c) 6.25 gram****(d) None****Ans.** (a)**Solution.**

We know that

where;

W = Mass of Na_{2}CO_{3} in gram

M = Molecular mass of Na_{2}CO_{3} in grams = 106

V = Volume of solution in litres = 250/1000 = 0.25

Molarity = 1/10

Hence, = 1/10

or**= 2.65 gram****Example ****5. The freezing point of a 0.08 molal aqueous solution of NaHSO _{4} is -0.372Â°C. The dissociation constant for the reaction**

Î”T

This means that total molal conc. of all particles is 0.2.

NaHSO

0.08 0.08

HSO

(0.08 - x) x x

The net particle concentration

0.087 = (0.08 - x) + x + x = 0.2

or x = 0.04

H

wt. of H

Vol of solution = 100ml

Moles of H

Weight of water = 198 - 95 = 103 g

= 9.412

Hence molality of H

Given H

wt. of H

Volume of solution = 100ml

weight of solution = 100 Ã— 1.84 gm

= 184 gm

wt. of water = 184 - 93 = 91 gm

= 10.42

**Fig: Acetic acid structure**

Wt. of CH_{3}COOH dissolved = 5g

Eq. of CH_{3}COOH dissolved = 5/60

Volume of ethanol = 1 litre = 1000ml.

Weight of ethanol = 1000 Ã— 0.789 = 789g

Molality of solution

= 0.1056**Example ****9.** **Calculate the molarity and normality of a solution containing 0.5 gm of NaOH dissolved in 500 ml. solution-****(a) 0.0025 M, 0.025 N ****(b) 0.025 M, 0.025 N****(c) 0.25 M, 0.25 N ****(d) 0.025 M, 0.0025 N** ** ****Ans. **(b)**Solution.**

Wt. of NaOH dissolved = 0.5 gm

Vol. of NaOH solution = 500 ml**Calculation of molarity **

0.5 g of NaOH = 0.5/40 moles of NaOH

[âˆ´ Mol. wt of NaOH = 40]

= 0.0125 moles

Thus 500 ml of the solution contain NaOH = 0.0125 moles

1000 ml of the solution contain

= 0.025 M

Hence molarity of the solution =** 0.025 M****Calculation of normality **

Since NaOH is monoacidic;

Eq. wt. of NaOH = Mol. wt. of NaOH = 40

0.5 gm of NaOH = 0.5/40 gm equivalents = 0.0125 gm equivalents

Thus 500 ml of the solution contain NaOH = 0.0125 gm equ.

1000 ml of the solution contain

Hence normality of the solution = **0.025 N****Example ****10.** **Calculate the molality and mole fraction of the solute in aqueous solution containing 3.0 gm of urea per 250 gm of water (Mol. wt. of urea = 60).****(a) 0.2 m, 0.00357 ****(b) 0.4 m, 0.00357 ****(c) 0.5 m, 0.00357 ****(d) 0.7m, 0.00357****Ans. **(a)**Solution.**

Wt. of solute (urea) dissolved = 3.0 gm

Wt. of the solvent (water) = 250 gm

Mol. wt. of the solute = 60

3.0 gm of the solute = 3.0/60 moles = 0.05 moles

Thus 250 gm of the solvent contain = 0.05 moles of solute

1000 gm of the solvent contain

= 0.2 moles

Hence molality of the solution = 0.2 m**In short,**

Molality = No. of moles of solute/1000 g of solvent

= **0.2 m****Calculation of mole fraction **

3.0 gm of solute = 3/60 moles = 0.05 moles

250 gm of water = 250/18 moles = 13.94 moles

Mole fraction of the solute

= **0.00357 ****Example ****11.** **A solution has 25% of water, 25% ethanol and 50% acetic acid by mass. Calculate the mole fraction of each component.****(a) 0.50, 0.3, 0.19 ****(b) 0.19, 0.3, 0.50 ****(c) 0.3, 0.19, 0.50 ****(d) 0.50, 0.19, 0.3****Ans. **(d)**Solution.**

Since 18 g of water = 1 mole

25 g of water = 25/18 = 1.38 mole

Similarly, 46 g of ethanol = 1 mole

25 g of ethanol = 25/46 = 0.55 moles

Again, 60 g of acetic acid = 1 mole

50 g of acetic acid = 50/60 = 0.83 mole

Mole fraction of water

= **0.50**

Similarly, Mole fraction of ethanol

=** 0.19**

Mole fraction of acetic acid

= **0.3****Example ****12. 15 gram of methyl alcohol is dissolved in 35 gram of water. What is the mass percentage of methyl alcohol in solution?****(a) 30% ****(b) 50% ****(c) 70% ****(d) 75%****Ans. **(a)**Solution.**

Total mass of solution = (15 35) gram = 50 gram

mass percentage of methyl alcohol

= **30% ****Example ****13. Calculate the masses of cane sugar and water required to prepare 250 gram of 25% cane sugar solution?****(a) 187.5 gram, 62.5 gram ****(b) 62.5 gram, 187.5 gram****(c) 162.5 gram, 87.5 gram ****(d) None of these****Ans. **(b)**Solution.**

Mass percentage of sugar = 25

We know that

Mass percentage

or Mass of cane sugar

= **62.5 gram**

Mass of water = (250 - 62.5) = **187.5 gram **

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

255 videos|306 docs|201 tests

### Raoult's Law

- Doc | 5 pages
### Vapour Solutions and Types of Solutions

- Doc | 7 pages
### Test: Solubility

- Test | 10 ques | 15 min
### Fun Video: What are the Solutions ?

- Video | 08:21 min
### Test: Solutions 1 - From Past 28 Years Questions

- Test | 17 ques | 35 min
### NCERT Textbook - Solutions

- Doc | 30 pages

- Previous year Questions (2016-20) - Solutions
- Doc | 12 pages
- Types of Solutions
- Video | 02:05 min