11th Class Chemistry Chapter one Short Questions with Answers Solved

Question: 1

The atomic mass may be in fractions. Why?

Answer: 1

1-21

Most of the elements have isotopes. Isotopes have fractional atomic masses. The average of atomic masses of isotopes will be definitely in fractions. Moreover, the monoisotopic elements have also fractional atomic, masses.

Question: 2

Answer: 2

2-21

Question: 3

What do you mean by molecular ions?

Answer: 3

3-21

Question: 4

What is atomic mass unit? Give its values in grams.

Answer: 4

4-21

It is a unit of mass used for atoms and molecules and is equal to the 1/12 of the mass of an atom of carbon —12. It is obtained by dividing the unity by Avogadro's number (6.02 x 10²³).

Question: 5

Why positively charged ions of isotopes are passed through magnetic field in the mass spectrometer?

Answer: 5

5-21

The positively charged ions bend perpendicular to the joining lines of the two poles, when passed through the magnetic field. In this way, magnetic field gives these ions semicircular path, scatters them on the basis of m/e values and compels them to fall on the electrometer. Electrometer develops the electric current. The strength of the torrent gives the relative of ions of definite m/v value.

Question: 6

Answer: 6

6-21

Question: 7

Define Avogadro's number. How does it relate to the masses of chemical substances?

Answer: 7

7-21

It is the number of atoms, molecules or ions in one gram mole of an element, compound and ion. One gram mole of the substance is the atomic mass, molar mass or ionic mass taken in grams. It means that the number of the species is related with the masses of the species.

Question: 8

Define empirical formula and how it is related to molecular formula?

Answer: 8

8-21

Question: 9

Molecular formula is multiple of empirical formula. Give an example.

Answer: 9

9-21

The empirical formula is the simplest whole number ratio of atoms of different elements present in a compound. Molecular formula tells us the actual number of atoms of each element in a molecule. Hence the empirical formula has to be multiplied with a suitable digit to get the molecular formula.

Question: 10

180 grams of glucose and 342 grams of sucrose have same number of molecules, but different number of atoms present in them. Justify it.

Answer: 10

10-21

180 grams of glucose (C₆H₁₂O₆) and 342 grams of sucrose (C₁₂H₂₂O₁₁) are one mole of each. One mole of various substances contain equal number of molecules i.e., 6.02 x 10²³. One molecule of C₅H₁₂O₆ has 24 atoms. The total number of atoms of glucose in one mole is 24 x 6.02 x 10²³.one molecule of C12H22011 has 45 atoms and the number of atom in 1 mole is 45 x 6.022 x 10²³. It means that one mole of both glucose and sucrose will have different number of atoms.

Question: 11

2 grams of H₂, 16g of CH₄ and 44g of CO₂ , occupy separately the volumes of 22.414 dm² at STP although the sizes and masses of molecules of three gases are very different from each other.

Answer: 11

11-21

One mole of an ideal gas at S.T.P. occupies a volume of 22.414 dm³. Sizes and masses of molecules of different gases do not affect the volume. Normally it is known that in the gaseous state, the distance between the molecules is 300 times greater than their diameter. Therefore two grams of H₂ , 16g of CH₄ and 44g of CO₂ (1 mole of each gas) separately occupy a volume of 22.4 dm³. This is called molar volume (V_m).

Question: 12

Calculate the mass in grams of 10^-2 moles of water.

Answer: 12

12-21

Question: 13

Answer: 13

13-21

Question: 14

What is limiting reactant? How it helps to control the reaction?

Answer: 14

14-21

A limiting reactants is that one which is in lesser quantity and it is consumed earlier. Whenever, it is consumed then the further formation of the product stops, although the excess reagent is lying in the vessel. If the limiting reagent is not available to the excess quantity then product cannot be formed further.

Question: 15

N₂ and CO have same number of electrons, protons and neutrons. Justify.

Answer: 15

15-21

No. of electrons in N₂ = 7 + 7 = 14, number of protons in N₂ = 7 + 7 = 14 and number of neutrons (14 - 7) = 7 + 7 = 14.

In CO, number of electrons in C = 6

Number of electrons in 0 = 8

Total number of electrons = 6 + 8 = 14

Number of neutrons in C = 6

Number of protons in 0 = 8

Total number of protons = 6 + 8 = 14

Question: 16

Define Solubility product and write down its application?

Answer: 16

16-21

Solubility product: The product of molar concentration of oppositely charged ions , in equilibrium with its solid salt in its saturated solution at a given temperature is called solubility product. Its is denoted by K_SP

Application of K_Sp:

Determination of K_SP From solubility.
Effect of common ion on solubility.

Question: 17

Prove that one mole of each 14₂, CO₂ and H₂ contain equal number of molecules.

Answer: 17

17-21

This is according to Avogadro's law that one mole of a substance has 6.02 10²³ molecules in it. So, 28 g of N₂ , 44g of CO₂ and 2 g of H₂ have 6.02 10²³ molecules in each.

Question: 18

How can the efficiency of a chemical reaction be expressed?

Answer: 18

18-21

Question: 19

What is limiting reactant? How it helps to control the reaction?

Answer: 19

19-21

A limiting reactants is that one which is in lesser quantity and it is consumed earlier. Whenever, it is consumed then the further formation of the product stops, although the excess reagent is lying in the vessel. If the limiting reagent is not available to the excess quantity then product cannot be formed further.

Question: 20

Distinguish between actual and theoretical yield.

Answer: 20

20-21

Question: 21

What the experimental yield is mostly less than the theoretical yield?

Answer: 21

21-21

. Experimental yield is mostly less than theoretical yield due to the following reasons.

(1) Mechanical loss of products due to:

(a) . Filtration

(b) Separation by distillation

(c) Separating by separating funnel

(d) Washing

(e) Drying

(f) Crystallization

(2)Side reaction.

(3)Reversibility of reaction.

Chemistry Chapters

11th Class Chapter one Chemistry Short Questions

0

Perparation Level