Subtopics of NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry
- Importance of Chemistry
- Nature of Matter
- Properties of Matter and Their Measurement
- The International System of Units (SI)
- Mass and Weight
- Uncertainty in Measurement
- Scientific Notation
- Significant Figures
- Dimensional Analysis
- Laws of Chemical Combinations
- Law of Conservation Of Mass
- Law of Definite Proportions
- Law of Multiple Proportions
- Gay Lussac’s Law of Gaseous Volumes
- Avogadro’s Law
- Dalton’s Atomic Theory
- Atomic and Molecular Masses
- Atomic Mass
- Average Atomic Mass
- Molecular Mass
- Formula Mass
- Mole Concept and Molar Masses
- Percentage Composition
- Empirical Formula for Molecular Formula
- Stoichiometry and Stoichiometric Calculations
- Limiting Reagent
- Reactions in Solutions
Access the answers of NCERT Solutions for Class 11 Chemistry Chapter 1
Exercise
Q1. Calculate the molar mass of the following:
(i)
Mass percent of the sulphur element:
=
Q4. Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
Ans.
(i) 1 mole of carbon is burnt in air.
(iii) 2 moles of carbon are burnt in 16 g of O2.
Here again, dioxygen is the limiting reactant. 16g of dioxygen can combine only with 0.5mol of carbon. CO2 produced again is equal to 22g.
Q5. Calculate the mass of sodium acetate
Q7. How much copper can be obtained from 100 g of copper sulphate (CuSO4)?
Ans.
1 mole of
Q8. Determine the molecular formula of an oxide of iron, in which the mass percent of iron and oxygen are 69.9 and 30.1, respectively.
Ans.
Here,
Mass percent of Fe = 69.9%
Mass percent of O = 30.1%
No. of moles of Fe present in oxide
=
Ans.
Fractional Abundance of 35Cl = 0.7577 and Molar mass = 34.9689
Fractional Abundance of 37Cl = 0.2423 and Molar mass = 36.9659
Average Atomic mass = (0.7577 x 34.9689)amu + (0.2423 x 36.9659)
= 26.4959 + 8.9568 = 35.4527
Q10. In three moles of ethane (C2H6), calculate the following:
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atom
(iii) Number of molecules of ethane
Ans.
(i) 1 mole of
or V1 = 0.02522L = 25.22mL
Q13. Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below:
1Pa = 1N m–2
If mass of air at sea level is 1034 g cm–2, calculate the pressure in pascal
Ans.
Pressure is the force (i.e., weight) acting per unit area
But weight = mg
∴ Pressure = Weight per unit area
Ans.
Significant figures are the meaningful digits which are known with certainty. Significant figures indicate uncertainty in experimented value.
e.g.: The result of the experiment is 15.6 mL in that case 15 is certain and 6 is uncertain. The total significant figures are 3.
Therefore, “the total number of digits in a number with the last digit that shows the uncertainty of the result is known as significant figures.”
Q17. A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).
(i) Express this in per cent by mass.
(ii) Determine the molality of chloroform in the water sample.
Ans.
(i) 1 ppm = 1 part out of 1 million parts.
Mass percent of 15 ppm chloroform in H2O
=
Q23. In a reaction
A + B2 → AB2
Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
Ans.
Limiting reagent:
It determines the extent of a reaction. It is the first to get consumed during a reaction, thus causes the reaction to stop and limits the amount of product formed.
(i) 300 atoms of A + 200 molecules of B
1 atom of A reacts with 1 molecule of B. Similarly, 200 atoms of A reacts with 200 molecules of B, so 100 atoms of A are unused. Hence, B is the limiting reagent.
(ii) 2 mol A + 3 mol B
1 mole of A reacts with 1 mole of B. Similarly, 2 moles of A reacts with 2 moles of B, so 1 mole of B is unused. Hence, A is the limiting reagent.
(iii) 100 atoms of A + 100 molecules of Y
1 atom of A reacts with 1 molecule of Y. Similarly, 100 atoms of A reacts with 100 molecules of Y. Hence, it is a stoichiometric mixture where there is no limiting reagent.
(iv) 5 mol A + 2.5 mol B
1 mole of A reacts with 1 mole of B. Similarly 2.5 moles of A reacts with 2.5 moles of B, so 2.5 moles of A is unused. Hence, B is the limiting reagent.
(v) 2.5 mol A + 5 mol B
1 mole of A reacts with 1 mole of B. Similarly, 2.5 moles of A reacts with 2.5 moles of B, so 2.5 moles of B is unused. Hence, A is the limiting reagent.
Q24. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
N2 (g) + H2(g)→ 2NH3 (g)
(i) Calculate the mass of
2 volumes of dihydrogen react with 1 volume of dioxygen to produce two volumes of water vapour.
Hence, 10 volumes of dihydrogen will react with five volumes of dioxygen to produce 10 volumes of water vapour.
Q27. Convert the following into basic units:
(i) 28.7 pm
(ii) 15.15 pm
(iii) 25365 mg
Ans.
(i) 28.7 pm
1 pm =
Therefore, 1 g of Li (s) will have the largest no. of atoms.
Q29. Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).
Ans.
Mole fraction of
Therefore, no. of significant numbers in the answer
= No. of significant numbers in 0.112
= 3
(ii) 5 × 5.364
Least precise number = 5.364
Therefore, no. of significant numbers in the answer
= No. of significant numbers in 5.364
= 4
(iii) 0.0125 + 0.7864 + 0.0215
As the least no. of decimal place in each term is 4. Hence, the no. of significant numbers in the answer is also 4.
Q32. Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:
Q35. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3 (s) + 2 HCl (aq) → CaCl2(aq) + CO2 (g) + H2O(l)
What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?
Ans.
0.75 M of HCl
≡ 0.75 mol of HCl are present in 1 L of water
≡ [(0.75 mol) × (36.5 g mol–1 )] HCl is present in 1 L of water
≡ 27.375 g of HCl is present in 1 L of water
Thus, 1000 mL of solution contains 27.375 g of HCl
Therefore, amt of HCl present in 25 mL of solution
=
Q36. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction:
4 HCl (aq) + MnO2(s) → 2H2O (l) + MnCl2(aq) + Cl2 (g)
How many grams of HCl react with 5.0 g of manganese dioxide?
Ans.
1 mole of
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