**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 H_{2}SO_{4 }= 2 + 32 + 64

Moles of H_{2}SO_{4}= 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 Co_{3}^{-2 }ions are present in it?**

**Solution:**

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

1 Molecule of CaCo_{3 }Gives 1 Ca^{+2 }and 1 Co_{3}^{-2 }ions

Given Mass = 40g

Moles =?

Molar mass of CaCo_{3 }= 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 × 10^{23})

= 2.4 × 10^{23}

Number of Co_{3}^{-2 }ions = 1 (moles × N_{A})

= 1 (0.4 × 6.02 × 10^{23})

= 204 × 10^{23}

40g of calcium carbonate has 2.4 × 10^{23 }Ca^{+2 }ions and 2.4 × 10^{23} Co_{3}^{-2} ions. (**Answer**)

**Question no: 3**

**If you have 6.02 × 10**^{23 }ions of aluminum how many sulphate ions will be required to prepare Al_{2}(SO_{4})_{3.}

**Solution:**

** Al**_{2}(SO_{4})_{3 } —————-> 2Al^{+ } + 3SO_{4}^{-2}

1 molecule of Al_{2}(SO_{4})_{3 }gives 2Al^{+3} ions and 3SO_{4}^{-2 }ions.

- If we have 1 mole of Al
_{2}(SO_{4})_{3 }then

Number of Al^{+3 }Ions are = 2(6.02 × 10^{23})

Number of SO_{4}^{-2 }ions are = 3 (6.02 × 10^{23})

- If we have ½ mole of Al
_{2}(SO_{4})_{3 }then

Number of Al^{+3 }ions are = 1/2 × 2 (6.02 × 10^{23})

= 6.02 × 10^{23}

Number of SO_{4}^{-3 }ions are = 1/2 × 3 (6.02 × 10^{23})

= 9.03 × 10^{23}

So if we have 6.02 × 10^{23 }ions of aluminum then 9.03 × 10^{23 }Sulphate ions will be required. (**Answer**)

**Question no: 4**

**Calculate number of molecules in **

**A: 16g of H _{2}CO_{3 }**

**B: 20g of HNO _{3 }**

**C: 30g of C _{6}H_{12}O_{6.}**

**Solution: **

**A) ** **16g of H _{2}CO_{3}**

Given mass = 16g

Molar mass = 2+12+48 = 62g

Moles of H_{2}CO_{3 }= ?

**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** ***×*** **10^{23}

^{ } = **1.56 ***×*** 10 ^{23}**

**B) ****20g of HNO _{3}**

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 × 10^{23}

^{ }= 1.86 × 10^{23}

**C) ****30g C _{2}H_{12}O_{6}**

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 × 10^{23}

^{ }= **4.57 × 10 ^{23 }**