11. Calculate the moles of chlorine atoms in 0.882g C2H4Cl2?
| Given mass (m) | =0.822g |
| Number of moles of Cl atom (n) | =? |
| The molar mass of C₂H4Cl₂ (M) | = 99 g mol-1 |
| No. of moles of C₂H4Cl₂ | = 0.822/99 =0.00830=8.30×10-3 |
| One mole of C₂H4Cl₂ | = 2 moles of Cl atoms |
| 8.30×10-3moles of C₂H4Cl₂ | = 2x 8.30×10-3 = 0.017 moles of Cl atoms. |
12. N₂ and CO have the same number of electrons, protons and neutrons. Explain with reason.
The number of electrons, protons and neutrons in nitrogen and carbon monoxide.
| N2 | CO | |
| – | N+N | C+O |
| Electrons | 7+7= 14 | 6+8= 14 |
| Protons | 7+7= 14 | 6+8= 14 |
| Neutrons | 7+7= 14 | 6+8= 14 |
So, N2 and CO have the same no. of electrons, protons and neutrons.
13. Differentiate between homoatomic and heteroatomic molecules with examples?
Homoatomic Molecules:
The molecules which consist of the same type of atoms are called homoatomic molecules. e.g. N₂, Cl2, O3 etc.
Heteroatomic Molecules:
The molecules which consist of different types of atoms are called heteroatomic molecules. e.g. HCl, H₂SO4 etc.
14. Calculate the mass in grams of 10-3 moles of H2O. The molecular mass of H2O is 18.
| No. of moles of H₂O(n) | = 10-3 moles |
| Molar mass of H₂O (M) | = 18g/mole |
| (Formula) n | = m/M |
| 10-3 | = m/18 |
| M | = 10-3 x18 |
| = 0.018g |
15. 100cm³ of NH3 gas and 100cm³ of H₂ gas at STP contain equal number of molecules. Justify it?
Different gasses have different molecular masses. The size of each gaseous atom is different from each other but one mole of each gas has 6.02×1023 particles. So, only empty spaces between molecules decrease by an increase in size. So, at S.T.P both NH3 and H₂ have the same number of molecules.
16. Molecular formula is an nth multiple of the empirical formula. Explain with examples.
Formula: (Molecular formula = n x Empirical formula)
Where “n” is a simple integer. Those compounds whose molecular and empirical formulas are the same have the “n” value unity. The formula may also be written as
(n = Molecular formula mass / Empirical formula mass)
For example, the Molecular formula of glucose is C6H12O6 and the empirical formula is
CH2O. Now according to the above formula, n= 180/30 => 6
17. By using a balanced chemical equation. What type of relationships can be studied?
The following types of relationships can be studied with the help of a balanced chemical equation.
- Mass-Mass relationship
- Mass-Mole relationship
- Mass-Volume relationship
18. Actual yield is usually less than the theoretical yield. Why?
The actual yield of a chemical reaction is always less than its theoretical yield due to the following reasons:
- Formation of by-products.
- Due to reversible reaction.
- Due to mechanical loss.
19. Calculate the number of gram atoms (moles) of Na when its mass is 0.1kg? The atomic mass of Na is 23g/ mol
| Gram atoms of Na | =? |
| Given mass of Na | = 0.1 kg = 100 g |
| Atomic mass of Na | =23 g/mole |
| (Formula) n | = mass/Molar mass |
| = 100/23 | |
| = 4.34 moles |
20. How many molecules of H2O are present when its amount is 0.25 moles?
| No. of H₂O molecules (N) | =? |
| No. of moles of H2O(n) | = 0.25 mole |
| Avogadro’s (NA) | = 6.02×1023 |
| Formula (No of molecules) N | = n x NA |
| = 0.25 x 6.02×1023 | |
| No. of H2O molecules | = 1.505×1023 |
