Solved Numerical of 9th Class Chemistry Chapter 1

 

Solved Numerical of 9th Class Chemistry Chapter 1 

Numerical 9th Class Solved Numerical of 9th Class Chemistry Chapter 1

Numerical 9th Class

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 490g 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:

CaCo3      ———–>    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 × NA)

= 1 (0.4 × 6.02 × 1023)

= 2.4 × 1023

Number of Co3-2 ions = 1 (moles × NA)

= 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) —————->  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 HNO

C: 30g of C6H12O6.

Solution:

A)     16g of H2CO3

Given mass = 16g

Molar mass = 2+12+48 = 62g

Moles of H2CO= ?

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 ×  NA

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 × NA

                                = 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 × NA

                                                                        = 0.76 × 6.02 × 1023

                                                                        = 4.57 × 1023 

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