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Chemistry
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(Chapter 1)(Some Basic Concepts of Chemistry)(Class 11)
Easy
Question 1:
State the number of significant figures in (i) 0.436 and (ii) 802.42
Answer 1:
(i) 0.436 has three significant figures (ii) 802.42 has five significant figures.
Question 2:
What do you mean by Mole fraction?
Answer 2:
Mole fraction is the ratio of number of moles of one component to the total
number of moles (solute & solvents) present in the solution. It is expressed as ‘x’.
Question 3:
Calculate the molecular mass of H2SO4.
Answer 3:
Molecular Mass of H2SO4 = 2 × 1 + 32 + 4 × 16 = 2 + 32 + 64 = 98u
Question 4:
What is the SI unit of mass? How is it de – fined?
Answer 4:
The S.I. unit of mass is Kilogram. Mass of a substance is the amount of matter
present in it.
Question 5:
What is limiting reagent?
Answer 5:
The reactant which gets consumed first or limits the amount of product formed is
known as limiting reagent.
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Chemistry
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(Chapter 1)(Some Basic Concepts of Chemistry)(Class 11)
Average
Question 1:
What is the value of Avogadro number?
Answer 1:
6.022 × 1023
Question 2:
What is the law called which deals with the ratios of the volumes of the gaseous
reactants and products of the volumes of the gaseous reactants and products?
Answer 2:
Gay Lussac’s law of gaseous volumes.
Question 3:
How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?
Answer 3:
Molar mass of Na2CO3 = 2 × 23 + 12 × 3 × 16 = 106 g/mol.
0.50 mol Na2CO3 means 0.50 × 106 = 53 g.
0.50 mol Na2CO3 means 0.50 mol i.e. 53 g of Na2CO3 are present in 1 L of the
solution.
Question 4:
What is the percentage of carbon, hydrogen and oxygen in ethanol?
Answer 4:
Molecular formula of ethanol is C2H5OH Molar mass of ethanol is
= 2 × 12 + 5 × 1 + 16 + 1
= 24 + 5 + 17 = 46 g
Mass percent of carbon =
24
46
Mass percent of hydrogen =
Mass percent of oxygen =
16
46
× 100 = 52.17%
6
46
× 100 = 13.1%
× 100 = 34.7%
Question 5:
Classify the following as pure substances or mixture.
(a) ethyl alcohol (b) oxygen (c) carbon (d) distilled water(e) blood (f) steel
Answer 5:
Pure substance – ethyl alcohol, oxygen, carbon, distilled water mixture – blood,
steel
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Chemistry
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(Chapter 1)(Some Basic Concepts of Chemistry)(Class 11)
Difficult
Question 1:
What is the difference between 160 cm and 160.0 cm?
Answer 1:
160 has three significant figures while 160.0 has four significant figures.
Hence160.0 represents greater accuracy.
Question 2:
In the combustion of methane in air, what is the limiting reagent & why?
Answer 2:
Methane is the limiting reagent because the other reactant is oxygen of air which
is always present in excess. Thus the amounts of carbon dioxide & water formed
will depend upon the amount of methane burnt.
Question 3:
Which aqueous solution has higher concentration, 1 molar or 1 molal solution of
the same solute? Give reason.
Answer 3:
1 molar aqueous solution has higher concentration than 1 molal solution. A molar
solution contains one mole of solute in one litre of solution while one molal
solution contains one mole of solute in 1000 g of solvent. If density of water is 1,
then one mole of solute is present in 1000 mL of water in 1 molal solution while
one mole of solute is present in less than 1000mL of water in 1 molar solution
(1000 mL sol = amount of solute + amount of solvent). Thus 1 molar solution is
more concentrated.
Question 4:
Calculate the molarity of water if its density is 1000 kg/m3.
Answer 4:
Molarity of water means the number of moles of water in 1 litre of water
1L of water = 1000 cm3 = 1000 g (1000 kg/m3 = 1g/cm3)
1000 g of water =
1000
18
= 55.56 moles
Molarity = 55.56 M
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