Solved Numerical of 9th Class Chemistry Chapter 1

Solved Numerical of 9th Class Chemistry Chapter 1 

Students are advised to note the numerical carefully. This is the Latest Numerical according to the new pattern of boards.

Question no: 1

Sulphuric acid is the king of chemicals if you need 5 moles of Sulphuric acid for a reaction. How many grams of it will you weigh?

Solution:

Molar mass of H₂SO₄ = 2 + 32 + 64

Moles of H₂SO₄= 5

Mass/Weight=?

Formula:

Unknown mass = moles * molar mass

= 5 * 98g

We will weigh 490 g of sulphuric acid to get 5 moles  (Answer)

Question no: 2

Calcium Carbonate is insoluble in water. If you have 40g of it how many Ca⁺² and Co₃^-2 ions are present in it?

Solution:

CaCo_{3      ———–>}    Ca⁺²   +  Co₃^-2

1 Molecule of CaCo₃ Gives 1 Ca⁺² and 1 Co₃^-2 ions

Given Mass = 40g

Moles =?

Molar mass of CaCo₃ = 100g

Formula

Moles = Given Mass/Molar Mass

Moles =  40/100 = 0.4

Number of Ca⁺² ions = 1 (moles × N_A)

= 1 (0.4 × 6.02 × 10²³)

= 2.4 × 10²³

Number of Co₃^-2 ions = 1 (moles × N_A)

= 1 (0.4 × 6.02 × 10²³)

= 204 × 10²³

40g of calcium carbonate has 2.4 × 10²³  Ca⁺² ions and 2.4 × 10²³ Co₃^-2 ions. (Answer)

Question no: 3

If you have 6.02 × 10²³ ions of aluminum how many sulphate ions will be required to prepare Al₂(SO₄)_3.

Solution:

  Al₂(SO₄)₃   —————->  2Al⁺  + 3SO₄^-2

1 molecule of Al₂(SO₄)₃ gives 2Al⁺³ ions and 3SO₄^-2 ions.

• If we have 1 mole of Al₂(SO₄)₃ then

Number of Al⁺³ Ions are = 2(6.02 × 10²³)

Number of SO₄^-2 ions are = 3 (6.02 × 10²³)

• If we have ½ mole of Al₂(SO₄)₃ then

Number of Al⁺³ ions are = 1/2 × 2 (6.02 × 10²³)

= 6.02 × 10²³

Number of SO₄^-3 ions are = 1/2 × 3 (6.02 × 10²³)

= 9.03 × 10²³

So if we have 6.02 × 10²³ ions of aluminum then 9.03 × 10²³ Sulphate ions will be required. (Answer)

Question no: 4

Calculate number of molecules in

A: 16g of H₂CO₃   

B: 20g of HNO₃ 

C: 30g of C₆H₁₂O_6.

Solution:

A)     16g of H₂CO₃

Given mass = 16g

Molar mass = 2+12+48 = 62g

Moles of H₂CO₃  = ?

Formula:-

Moles = given mass/Molar mass

Putting the value of mass and molar mass in above formula you get

Moles: 16/62 = 0.26

Number of moles = ?

Formula:-

 Number of moles = Moles ×  N_A

By putting the value

Number of moles = 0.26 ×  6.02 ×  10²³

                                                   = 1.56 ×  10²³

B)      20g of HNO₃

Given mass = 20g

Molar mass = 1 + 14 + 48

= 63g

Moles = ?

Formula:-     Moles = given mass/ molar mass

By putting the value

Moles = 20/63 = 0.31

Number of moles = ?

Formula: – number of molecules = moles × N_A

                                = 0.31 × 6.02 × 10²³

                                                = 1.86 × 10²³

C)      30g C₂H₁₂O₆

Given mass = 30g

Molar mass = 72 + 12 + 96 = 180g

Moles = given mass/molar mass

= 30/180 = 0.76

Number of molecules = moles × N_A

                                                                        = 0.76 × 6.02 × 10²³

                                                                        = 4.57 × 10²³