Atoms And Molecules Class 9th Physics & Chemistry AP Board Solution

Class 9^th Physics & Chemistry AP Board Solution

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1. Draw the diagram to show the experimental setup for the law of conservation of mass.…
2. Explain the process and precautions in verifying law of conservation of mass.…
3. 15.9g. of copper sulphate and 10.6g of sodium carbonate react together to give 14.2g of…
4. Carbon dioxide is added to 112g of calcium oxide. The product formed is 200g of calcium…
5. 0.24g sample of compound of oxygen and boron was found by analysis to contain 0.144g of…
6. In a class, a teacher asked to write the molecular formula of oxygen Shamita wrote the…
7. Imagine what would happen if we do not have standard symbols for elements?…
8. Mohith said "H2 differs from 2H". Justify.
9. Lakshmi gives a statement "CO and Co both represents element". Is it correct? State…
10. The formula of the water molecule is H2O. What information you get from this formula.…
11. How would you write 2 molecules of oxygen and 5 molecules of Nitrogen.…
12. The formula of a metal oxide is MO. Then write the formula of its chloride.…
13. Formula of calcium hydroxide is Ca (OH)2 and zinc phosphate is Zn3(PO4)2. Then write the…
14. Find out the chemical names and formulae for the following common household substances. a)…
15. 0.5 mole of N2 gas. Calculate the mass of the following
16. 0.5 mole of N atoms. Calculate the mass of the following
17. 3.011 × 10^23 number of N atoms. Calculate the mass of the following…
18. 6.022 × 10^23 number of N2 molecules. Calculate the mass of the following…
19. 46g of Na Calculate the number of particles in each of the following…
20. 8g of O2 Calculate the number of particles in each of the following…
21. 0.1 mole of hydrogen Calculate the number of particles in each of the following…
22. 12g of O2 gas. Convert into mole
23. 20g of water. Convert into mole
24. 22g of carbon dioxide. Convert into mole
25. Write the valencies of Fe in FeCl2 and FeCl3
26. Calculate the molar mass of Sulphuric acid (H2SO4) and glucose (C6H12O6)…
27. Which has more number of atoms - 100g of sodium or 100g of iron? Justify your answer.…
28. Complete the following table.
29. Fill the following table

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Question 1.

Draw the diagram to show the experimental setup for the law of conservation of mass.

Answer:

Step 1: Take a flask and take a mixture of sodium hydroxide solution and copper sulphate solution.

Step 2: Weight the flask and note it.

Step 3: Now mix the whole mixture thoroughly.

Step 4: Now, weight the flask.

We will observe that after mixing there is no change in the mass. This explains the law of conservation of mass.

Question 2.

Explain the process and precautions in verifying law of conservation of mass.

Answer:

Law of conservation of mass states that “Matter is neither created nor destroys in a chemical reaction. More simply, the mass of products is equal to the mass of reactants in a chemical reaction.

Question 3.

15.9g. of copper sulphate and 10.6g of sodium carbonate react together to give 14.2g of sodium sulphate and 12.3g of copper carbonate. Which law of chemical combination is obeyed? How?

Answer:

Law of conservation of mass is obeyed.

Explanation:

Total mass of reactants = 15.9g + 10.6g = 26.5g

Total mass of products = 14.2g + 12.3g = 26.5g

As there is no change in the mass of reactants and products, hence law of conservation is obeyed.

Question 4.

Carbon dioxide is added to 112g of calcium oxide. The product formed is 200g of calcium carbonate. Calculate the mass carbon dioxide used. Which law of chemical combination will govern your answer.

Answer:

The reaction taking place is:

Let the mass of CO₂ is xg.

According to the law of conservation of mass, the mass of reactants is equal to the mass of products.

Therefore,

⇒ x + 112g = 220g

⇒ x = 220g – 112g

⇒ x = 108g

Thus, the mass of carbon dioxide is 108g.

Question 5.

0.24g sample of compound of oxygen and boron was found by analysis to contain 0.144g of oxygen and 0.096g of boron. Calculate the percentage composition of the compound by weight.

Answer:

Given: Mass of compound = 0.24g

Mass of oxygen = 0.144g

Mass of boron = 0.096g

To calculate the percent of the composition of oxygen, we apply the formula:

⇒ 

⇒ % composition of oxygen = 0.6g × 100

⇒ % composition of oxygen = 60%

To calculate the percent of the composition of boron, we apply the formula:

⇒ 

⇒ % composition of boron = 0.4g × 100

⇒ % composition of boron = 40%

Total percentage composition of compound is 60% + 40% = 100%

Question 6.

In a class, a teacher asked to write the molecular formula of oxygen Shamita wrote the formula as O₂ and Priyanka as O. which one is correct? State the reason.

Answer:

Shamita wrote the correct formula of oxygen as O₂ because:

i. A molecule of oxygen has two atoms.

ii. It is diatomic in nature (atomicity = 2)

iii. Thus, writing oxygen as O₂ indicates two separate atoms of oxygen.

Note: Atomicity is the number of atoms constituting a molecule.

Question 7.

Imagine what would happen if we do not have standard symbols for elements?

Answer:

if we do not have standard symbols for elements, then it will take a lot of time to write the full name of the elements and compounds every time to describe a reaction.

Question 8.

Mohith said "H₂ differs from 2H". Justify.

Answer:

Question 9.

Lakshmi gives a statement "CO and Co both represents element". Is it correct? State reason.

Answer:

No, CO does not represent element whereas Co represents element because the first letter of the symbol is always the upper case and the second letter is always lower case.

In CO, the second letter is uppercase whereas in Co, the second letter is lowercase. Hence Co represents an element.

Question 10.

The formula of the water molecule is H₂O. What information you get from this formula.

Answer:

To get the information, first, we will see the elements present in a molecule of the compound.

Second, we will count the number of atoms of each element of that molecule.

In H₂O:

i. 2 atoms of hydrogen and one atom of oxygen are present.

ii. There are total three atoms present in a molecule of water.

Question 11.

How would you write 2 molecules of oxygen and 5 molecules of Nitrogen.

Answer:

For oxygen:

⇒ First, write the symbol of oxygen – 

⇒ Now write 2 as a subscript after O – 

For nitrogen:

⇒ First, write the symbol of nitrogen – 

⇒ Now write 5 as a subscript after N – 

Question 12.

The formula of a metal oxide is MO. Then write the formula of its chloride.

Answer:

The formula of its chloride is MOCl₂

Question 13.

Formula of calcium hydroxide is Ca (OH)₂ and zinc phosphate is Zn₃(PO₄)₂. Then write the formula to calcium phosphate.

Answer:

Valency of Ca = 2

Valency of PO₄ = 3

Now, apply the criss cross method

So the formula will be Ca₃(PO₄)₂

Question 14.

Find out the chemical names and formulae for the following common household substances.

a) common salt

b) baking soda

c) washing soda

d) vinegar

Answer:

a) Common salt

Chemical name – sodium chloride

Chemical formula – NaCl

b) baking soda

Chemical name – sodium bicarbonate

Chemical formula – NaHCO₃

c) washing soda

Chemical name – sodium carbonate

Chemical formula – Na₂CO₃

d) vinegar

Chemical name – acetic acid

Chemical formula – CH₃COOH

Question 15.

Calculate the mass of the following

0.5 mole of N₂ gas.

Answer:

0.5 mole of N₂ gas

First, we apply the formula:

⇒ Mass of N₂ = No. of moles × Molar mas of N₂

⇒ Mass of N₂ = 0.5 mole × 2(atomic mass of nitrogen)

⇒ Mass of N₂ = 0.5 × 2 × 14

⇒ Mass of N₂ = 14g

Thus, the mass of N₂ gas is 14g.

Question 16.

Calculate the mass of the following

0.5 mole of N atoms.

Answer:

0.5 mole of Natoms

First, we apply the formula:

⇒ Mass of Natoms = No. of moles × Molar mass of N

⇒ Mass of Natoms = 0.5 mole × (atomic mass of nitrogen)

⇒ Mass of Natoms = 0.5 × 14

⇒ Mass of N₂ = 7g

Thus, the mass of N atoms is 7g.

Question 17.

Calculate the mass of the following

3.011 × 10²³ number of N atoms.

Answer:

3.011 X 10²³ number of N atoms.

As we know that 1 mole = 6.022 X 10²³atoms

⇒ 

⇒ 1/2 mole = 0.5 mole

For 0.5 mole of Natoms

First, we apply the formula:

⇒ Mass of Natoms = No. of moles × Molar mass of N

⇒ Mass of Natoms = 0.5 mole × (atomic mass of nitrogen)

⇒ Mass of Natoms = 0.5 × 14

⇒ Mass of N atoms = 7g

Thus, mass of3.011 × 10²³ number of N atoms is 7g

Question 18.

Calculate the mass of the following

6.022 × 10²³ number of N₂ molecules.

Answer:

6.022 X 10²³ number of N₂ molecules.

As we know that 1 mole = 6.022 X 10²³atoms

For 1 mole of N₂molecules:

First, we apply the formula:

⇒ Mass of N₂ = No. of moles × Molar mass of N₂

⇒ Mass of N₂ = 1 mole × 2(atomic mass of nitrogen)

⇒ Mass of N₂ = 1mole × 28u

⇒ Mass of N₂ = 28g

Thus, mass of6.022 × 10²³ number of N₂ molecules is 28g.

Question 19.

Calculate the number of particles in each of the following

46g of Na

Answer:

The number of particles present in one mole of any substance is has a fixed value of 6.022 × 10²³. This number is called Avogadro constant(N_A).

Mass of Na = 46g

Molar mass of Na = 23u

Apply the formula given below:

Number of particles = number of moles × 6.022 × 10²³

⇒ 

⇒ 

⇒ Number of particles = 2 × 6.022 × 10²³

^⇒ Number of particles = 12.046 × 10²³

Thus, 46g of Na contains 12.046 × 10²³ particles.

Question 20.

Calculate the number of particles in each of the following

8g of O₂

Answer:

The number of particles present in one mole of any substance is has a fixed value of 6.022 × 10²³. This number is called Avogadro constant(N_A).

Mass of O₂ = 8g

Molar mass of O₂ = 32u

Apply the formula given below:

Number of particles = number of moles × 6.022 × 10²³

⇒ 

⇒ 

⇒ Number of particles = 0.25 × 6.022 × 10²³

^⇒ Number of particles = 1.5 × 10²³

Thus, 8g of O₂ contains 1.5 × 10²³ particles.

Question 21.

Calculate the number of particles in each of the following

0.1 mole of hydrogen

Answer:

The number of particles present in one mole of any substance is has a fixed value of 6.022 × 10²³. This number is called Avogadro constant(N_A).

Number of mole of hydrogen = 0.1 mole

Molar mass of H₂ = 2u

Apply the formula given below:

Number of particles = number of moles × 6.022 × 10²³

⇒ Number of particles = 0.1 mole × 6.022 × 10²³

^⇒ Number of particles = 6.022 × 10²⁴

Thus, 0.1 mole of hydrogen contains 6.022 × 10²⁴particles.

Question 22.

Convert into mole

12g of O₂ gas.

Answer:

Mass of O₂ gas = 12g

Molar mass of O₂ = 32u

Apply the formula:

⇒ 

⇒ No. of moles = 0.37

Thus, the 12g of O₂ gas is 0.37 mole

Question 23.

Convert into mole

20g of water.

Answer:

Mass of water(H₂O) = 20g

Molar mass of H₂O = 18u

Apply the formula:

⇒ 

⇒ No. of moles = 1.11

Thus, the 20g of water is 1.11 mole

Question 24.

Convert into mole

22g of carbon dioxide.

Answer:

Mass of CO₂ gas = 22g

Molar mass of CO₂ = 44u

Apply the formula:

⇒ 

⇒ No. of moles = 0.5

Thus, the 22g of CO₂ gas is 0.5 mole

Question 25.

Write the valencies of Fe in FeCl₂ and FeCl₃

Answer:

In FeCl₂, the valency of Fe is + 2

In FeCl₃, the valency of Fe is + 3

Question 26.

Calculate the molar mass of Sulphuric acid (H₂SO₄) and glucose (C₆H₁₂O₆)

Answer:

Molar mass of sulphuric acid (H₂SO₄):

⇒ 2(atomic mass of hydrogen) + (atomic mass of Sulphur) + 4(atomic mass of oxygen)

⇒ 2 × 1 + 32 + 4 × 16

⇒ 2 + 32 + 64

⇒ 98 u

Thus, the molar mass of H₂SO₄ is 98 u.

Molar mass of glucose (C₆H₁₂O₆):

⇒ 6(atomic mass of carbon) + 12(atomic mass of hydrogen) + 6(atomic mass of oxygen)

⇒ 6 × 12 + 12 × 1 + 6 × 16

⇒ 72 + 12 + 96

⇒ 108 u

Thus, the molar mass of C₆H₁₂O₆ is 108 u.

Question 27.

Which has more number of atoms - 100g of sodium or 100g of iron? Justify your answer. (atomic mass of sodium = 23u, atomic mass of iron = 56u)

Answer:

For sodium

Given: Mass of sodium = 100g

Molar mass of sodium = 23u

Number of atoms = number of moles × 6.022 × 10²³

⇒ 

⇒ 

⇒ Number of atoms = 4.34 × 6.022 × 10²³

^⇒ Number of atomss = 2.613 × 10²⁴

^⇒Thus, 100g of sodium contains 2.613 × 10²⁴ atoms.

For iron

Given: Mass of iron = 100g

Molar mass of iron = 55.8u

Number of atoms = number of moles × 6.022 × 10²³

⇒ 

⇒ 

⇒ Number of atoms = 1.79× 6.022 × 10²³

^⇒ Number of atoms = 1.07× 10²⁴

^⇒Thus, 100g of iron contains 1.07× 10²⁴atoms.

Therefore, 100g of sodium has more number of atoms.

Question 28.

Complete the following table.

Answer:

Question 29.

Fill the following table

Answer: