SOME BASIC CONCEPTS OF CHEMISTRY
Two Marks questions with answers
1. Give the two points of differences between homogeneous and heterogeneous mixtures.
Ans.
| Heterogeneous mixture | | | |||||
1. | Homogeneous | mixtures | 1. | Heterogeneous mixtures consist of | ||||
| have the same composition | | two or more parts (phases), which | |||||
| throughout the sample. | | have different compositions. | | ||||
| | | | 2. | These | mixtures | have | visible |
2. | The components of such | | boundaries of separation between the | |||||
| mixtures | cannot | be seen | | different constituents and can be seen | |||
| under | a | powerful | | with the naked eye | | | |
| microscope. | | | | | | |
2. Copper oxide obtained by heating copper carbonate or copper nitrate contains copper and oxygen in the same ration by mass. Which law is illustrated by this
observation? State the law.
Ans.Law of Definite Proportions This law states that: A chemical compound always consists of the same elements combined together in the same ratio, irrespective of the method of preparation or the source from where it is taken.
3. | Write the empirical formula of the following: | |||
| (a) N2O4 | (b) C6H12O6 (c) H2O | (d) H2O2 | |
Ans. (a)NO2 | (b) CH2O (c) H2O | (d) HO | ||
4. | Briefly explain the difference between precision and accuracy. | |||
Ans. Precision refers to the closeness of various measurements for the same quantity. However, accuracy is the agreement of a particular value to the true value of the result.
5. Define the law of multiple proportions. Explain it with one example. Ans.When two elements combine to form two or more compounds, then the different masses of one element, which combine with a fixed mass of the other, bear a simple ratio to one another. For example- carbon combines
with oxygen to form two compounds CO and CO2. | | |
Compound | CO | CO2 |
Mass of C | 12 | 12 |
Mass of O | 16 | 32 |
Masses of oxygen which combine with a fixed mass of carbon (12g) bear a simple ratio of 16:32 or 1:2.
6. Chlorine has two isotopes of atomic mass units 34.97 and 36.97. The relative abundance of the isotopes is 0.755 and 0.245 respectively. Find the average atomic mass of chlorine.
Ans. Average atomic mass = 34.97 x 0.755 +36.97 x 0.245 = 35.46 u
7. Calculate the percentage composition | of water. | ||
Ans. Mass % of an element = mass of that element in the compound × 100 | |||
molar mass of the compound | | | |
Molar mass of water = 18.02 g | | | |
Mass % of hydrogen = 2× 1.008 × 100 | | | |
18.02 | | | |
= 11.18 | | | |
Mass % of oxygen = 16.00 × 100 | = 88.79 | ||
| | | |
18.02 | | | |
(i) 208.91 (ii) 0.00456 (iii) 453 (iv) 0.346
Ans.
(i) 208.91 has five significant figures.
(ii) 0.00456 has three significant figures.
(iii) 453 has three significant figures.
(iv) 0.346 has three significant figures.
8. Express the results of the following calculations to the appropriate number of significant figures.
= 0.2615 x 10-4 = 0.3 x 10-4
9. How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different? Ans. Molar mass of Na2CO3= 2 x 23 +12 + 3 x 16 = 106 g / mol
0.50 molNa2CO3means 0.50 x 106 = 53 g
0.50 M Na2CO3 means 0.50 mol i.e. 53 g of Na2CO3 are present in I L of the solution.
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