4.
Numericals
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?
2.
Calcium carbonate is insoluble in water. If you have 40 g of it; how many
Ca2+ and CO32-ions are present in it?
3.
If you have 6.02 x 10 2 ions of aluminium; how many sulphate ions will
be required to prepare Al2(SO4)3?
Calculate the number of molecules in the following compounds:
a. 16 g of H2CO3 b. 20 g of HNO3 c. 30 g of C. H 2 O.
5. Calculate the number of ions in the following compounds:
a. 10 g of AlCl3 b. 30 g of BaCl2 c. 58 g of H2SO4(29)
6. What will be the mass of 2.05x1016 molecules of H2SO4
7. How
many atoms are required to prepare 60 g of HNO3?
8. How many ions of Nat and Cl" will be present in 30 g of NaCl?
9. How many molecules of HCI will be required to have 10 grams of it?
10. How many grams of Mg will have the same number of atoms as 6 grams
of Chave?
Answers
Answer:
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)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 C6H12O6.
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 × 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
Explanation: