NCERT Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry

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NCERT Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry

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Some Basic Concepts of Chemistry

Chapter: 1

1. Calculate the molar mass of the following: 

(i) H₂O

Ans: Molecular mass of H₂O

= 2(1.008 amu) + 16.00 amu

= 2.016 amu + 16.00 amu

= 16.016 amu.

(ii) CO₂

Ans: Molecular mass of CO₂ 

= 12.01 amu + 2 × 16.00 amu

= 12.01 amu + 32 amu

= 44.01 amu.

(iii) CH₄ 

Ans: Molecular mass of CH₄

= 12.1 amu + 4 × (1.008 amu)

= 12.1 amu + 4.032 amu

= 16.132 amu.

2. Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4). 

Ans: Mass % of an element

= Mass of that element in the compound / Molar mass of compound × 100 

Now, molar mass of  Na₂SO = 23 × 2 + 32 + 4 × 16 

= 46 +32 + 64 =142

Na = 23 × 2 / 142 × 100 

= 46 / 142 × 100

= 0.323 100

= 32.4 %

S = 32/142 × 100 

= 0.225 × 100 

= 22.5 %

O = 64/142 × 100 

= 0.4507 × 100

= 45.07 %

3. Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass. 

Ans: 

∴ Empirical formula – Fe₂O₃.

4. 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: The balanced equation for the combustion of carbon in dioxygen/air is 

(i) C + O₂ = CO₂

1 mole of carbon burnt in air will produce 44 grams of CO₂.

(ii) 1 mole of carbon is burnt in 16g of dioxygen.

16g of dioxygen corresponds to 16/32 = 0.5 moles. 

i.e., dioxygen is the limiting reactant. 

Hence, CO₂ produced= 0.5 × 44 = 22 g.

(iii) 2 mole of carbon are burnt 16g of dioxygen.

16g of of dioxygen corresponds to 16/32 = 0.5 mole. 

Here,dioxygen is the limiting reagent.

Hence, CO₂  corresponds = 0.5 × 44 = 22 g.

5. Calculate the mass of sodium acetate (CH₃COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–¹.

Ans: Molarity (M) = 0.375 M

Volume (V) = 500 mL = 0.500 L

Molar mass of sodium acetate (CH₃​COONa) = 82.0245 g/mol

Sodium acetate = 0.375/2 mole.

Moral mass of sodium acetate required = 0.375/2

Mole × 82.0245 g mol ^-1 = 15.380g. 

6. Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL–1 and the mass per cent of nitric acid in it being 69%. 

Ans: Mass percent of HNO₃ = 69%

Molar mass of nitric acid (HNO₃)

= 1 + 14 + 48= 63 g mol^-1

Moles of HNO3,n = 69/ HNO₃ 

= 69/63 = 1.095 

7. How much copper can be obtained from 100 g of copper sulphate (CuSO₄)? 

Ans: 1 mole of CuSO₄ contains 1 mole (1 g atom) of Cu

Copper (Cu): 63.55 g/mol

Sulphur (S): 32.07 g/mol

Oxygen (O): 16.00 g/mol × 4 = 64.00 g/mol

Molar mass of CuSO₄ 

= 63.5 + 32 + 4  ×  16

= 159.5 g mol-1

Thus, Cu that can be obtained from 159.5 g of CuSO₄ = 63.5g

Therefore Cu that can be obtained from 100g of CuSO₄

= 63.5/ 159.5 × 100g

= 39.81g.

8. Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively. 

Ans: Calculation of empirical Formula:

So, empirical formula of iron oxide is Fe₂O_3.

Empirical formula mass of Fe2O3 = 2  x 55.85 + 3 × 16.00 

= 111.7 + 48.00

= 159.7g mol-1

N = Molar mass/empirical formula mass 

= 159.8 / 159.7 = 1

Hence, Thus, the molecular formula of the oxide of iron is Fe₂O₃.

9. Calculate the atomic mass (average) of chlorine using the following data:

Ans: Fractional abundance of ³⁵Cl = 75.77% = 0.7577

Molar mass of ³⁵Cl = 34.9689g/ mol

Fractional abundance of ³⁷Cl =  24.23% = 0.2423

Molar mass of ³⁷Cl = 36.9659

∴ Average atomic mass of chlorine 

= (0.7577) (34.9689) + (0.2423) (36.9659)

=  0.7577 × 34.9689 = 26.4959

= 0.2423 × 36.9659 = 8.9566

= 26.4959 + 8.9566 = 35.4527.

10. In three moles of ethane (C₂H₆), calculate the following: 

(i) Number of moles of carbon atoms.

Ans: 1 mole of C₂H₆ contains 2 moles of carbon atoms.

Number of moles of carbon atoms = 3 × 2 = 6 mol

(ii) Number of moles of hydrogen atoms.

Ans: 1 mole of C₂H₆ contains 6 moles of hydrogen atoms.

Number of moles of hydrogen atoms = 3 × 6 = 18 mol

(iii) Number of molecules of ethane. 

Ans: 1 mole of C₂H₆ contains Avogadro,s no. i.e. 6.02 × 10 ²³ molecules.

Number of molecules of ethane = 3 × 6.023 × 10 ²³ 

= 18.069 × 10 ²³ molecules.

11. What is the concentration of sugar (C₁₂H₂₂O₁₁) in mol L–1 if its 20 g are dissolved in enough water to make a final volume up to 2L? 

Ans: Molar mass of sugar (C₁₂H₂₂O₁₁) 

= (12 × 12) + (22 × 1) + (11 ×16.00)

= 144 + 22+ 176

= 342.

No. of moles in 20g of sugar = Mass/molars 

= 20/342 g mol^-1 

= 0.0585 mol

Volume of solution = mole of solute/ volume of sol in L

= 0.0585 mol/ 2l

= 0.0293 mol L^-1.

12. If the density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its 0.25 M solution? 

Ans: Molar mass of methanol = Molar × Molar mass of methanol

= 0.625 × 32.042 g /mol

= 20.00 g

Calculation of mass methanol to volume using its density:

Volume = mass/ density = 20.00g / 0.793 kg × 1000 g/kg

= 0.025 × 1000 g/kg

= 25.22 ml.

13. 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: Force (F) = Mass (m) × Gravity (g)

Assuming = 9.81 m/s2

Mass per unit are, m/a = 1034 gm

Convert mass from g to kg: 1034g = 1.034 kg

Force = 1.034 × 9.81 m/ s² = 10.13734 N/m²

Convert the mass of air from g/cm² to kg/m

1034g/cm2 = 1034g/ cm² × 1kg / 1000 g × (100 cm/m)²

= 1034g × 0.001² 1

Pressure is the force (i.e, weight) acting per unit area

Weight = mg

Pressure = weight per unit area

= 1034g × 9.81 m/ s / cm² × 1kg /1000g × 100cm /1m × 1N / kg ms-2 × 1Pa / 1N m-2

= 1.01332 × 10⁵ pa.

14. What is the SI unit of mass? How is it defined?

Ans: Kilogram is used as SI unit of mass. It is represented by the symbol kg. It is defined as the mass of platinum-iridium  block stored at international bureau of weight and measures in france. The system of units, including unit definitions, keeps on changing with time. Whenever the accuracy of measurement of a particular unit was enhanced substantially by adopting new principles, member nations of metre treaty (signed in 1875), agreed to change the formal definition of that unit.

15. Match the following prefixes with their multiples:

Ans:  

16. What do you mean by significant figures?

Ans: The uncertainty in the experimental or the calculated values is indicated by mentioning the number of significant figures. Significant figures are meaningful digits which are known with certainty plus one which is estimated or uncertain. The uncertainty is indicated by writing the certain digits and the last uncertain digit.

17. 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.

Ans: 15 ppm means 15 parts in million (10⁶) parts

Percent by mass = Mass of chloroform/ mass of solution × 100

= 15/ 10⁶ × 100

= 1.5 × 10^-3 %

(ii) Determine the molality of chloroform in the water sample. 

Ans: Molar mass of chloroform (CHCL³)

= Mass of chloroform/ Molar mass of chloroform × (mass of solution – mass of chloroform) × 100

= 12 + 1 3 × 35.5 

= 118.5 g mol-1 

100g of the sample contain chloroform

= 1.5 × 10-3g 

1000g 91 kg of the sample will contain chloroform Molality = 15 × 10-2 / 118.65   mole

= 1.266 × 10^-4 mole 

18. Express the following in the scientific notation: 

(i) 0.0048

Ans: 0.0048 × 10^-3

(ii) 234,000 

Ans: 234,000 × 10⁵

(iii) 8008 

Ans: 8008 × 10³

(iv) 500.0 

Ans: 500.0 × 10²

(v) 6.0012 

Ans: 6.0012 × 10⁰

19. How many significant figures are present in the following? 

(i) 0.0025

Ans: Significant figure: 2

(ii) 208 

Ans: Significant figure: 3

(iii) 5005 

Ans: Significant figure: 4

(iv) 126,000 

Ans: Significant figure: 3

(v) 500.0 

Ans: Significant figure: 4

(vi) 2.0034 

Ans: Significant figure:5

20. Round up the following upto three significant figures: 

(i) 34.216

Ans: Three significant figures: 34.2

(ii) 10.4107 

Ans: Three significant figures: 10.4

(iii) 0.04597 

Ans: Three significant figures: 0.045

(iv) 2808 

Ans: Three significant figures: 280

21. The following data are obtained when dinitrogen and dioxygen react together to form different compounds:

(a) Which law of chemical combination is obeyed by the above experimental data? Give its statement. 

Ans: When the mass of dinitrogen is 28g, mass of dioxygen combined is 32, 64, 32, and 80g. The  corresponding ratio is 1:2:1:5. It is a simple whole number ratio. Hence the given data obey the low of multiple proportions.

Low of multiple proportions: When two elements combine to form two or more chemical compounds, then masses of one of the elements which  combine with a fixed mass of the other, bear a simple ratio to one another.

(b) Fill in the blanks in the following conversions: 

(i) 1 km = ____________ mm = ________ pm.

Ans: 1 km = 1 km × 1000m/1km × 100m/1m × 10mm /1cm

= 1000 × 100 × 10 

= 100000mm 

= 10⁶ mm 

1 km = 1 km × 1000m/1km × 1pm/ 10^-12 m

= 10¹⁵ pm 

(ii) 1 mg = ___________ kg = _____________ ng.

Ans: 1 mg = 1mg × 1g/1000g × 1mg/1000g

= 10^-6kg

1 mg = 1 mg × 1 /1000 mg × 1ng 10^-9g

= 10⁶ ng

mL = 1 cm³

(iii) 1 mL = ____________ L = _____________ dm3 

Ans: 1 ml = 1 ml × 1g/1000 mg 

= 10^-3 L

Ml = 1cm³

= 1 cm³ × 1 dm × 1 dm × 1 dm / 10 cm × 10 cm × 10 cm 

= 10^-3 dm³.

22. If the speed of light is 3.0 × 10⁸ m s–1, calculate the distance covered by light in 2.00 ns. 

Ans: Convert time from nanoseconds to seconds:

2.00 ns = 2.00 × 10⁻⁹ s

Distance = Speed × Time 

= 3.0 × 10⁸ m s–1 × 2.00 ns

= 3.0 × 10⁸ m s–1 × 2.00 ns × 10⁻⁹ s/ 1ns

= 6.00 × 10^-1 m s -1 = 0.600 m.

23. 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.

Ans: According to the given reaction, 1 atom of A reacts with 1 atom of B.

∴ 200 molecules of B will react with 200 atoms of A 100 atoms of A will remain unreacted Hence, B₂ is the limiting reagent.

(ii) 2 mol A + 3 mol B. 

Ans: According to the given reaction, 1 mol of A reacts with 1 mol of B.

∴ 2 mol of A will react with only 2 mol of B. As a result, 1 mol of B will not be consumed. Hence, Ais the limiting reagent. 

(iii) 100 atoms of A + 100 molecules of B. 

Ans: There is no limiting reagent.

(iv) 5 mol A + 2.5 mol B.

Ans: 2.5 mol of B will combine with only 2.5 mol of A. As a result, 2.5 mol of A will be left as such. Hence, B is the limiting reagent.

(v) 2.5 mol A + 5 mol B.

Ans: 2.5 mol of A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be left as such. Hence, A is the limiting reagent.

24. 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 ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 × 103 g of hydrogen. 

Ans: Mass of ammonia produced

28g of N₂ produces 34g of NH₃

1g of N₂ will produce = 34/28 g NH₃

2000 g of N₂ will produce NH₃ = 34/28 × 2000

= 2428.57 g of NH₃.

(ii) Will any of the two reactants remain unreacted? 

Ans: Yes,H₂ will remain unreacted. 

(iii) If yes, which one and what would be its mass? 

Ans: Mass left unreacted 

= 1000 g – 428.6 g 

= 571.4 g. 

25. How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different? 

Ans: 0.50 mol HCl means there is a total of 0.50 moles of HCl present, while 0.50 M HCl (M stands for molarity) indicates that there is a concentration of 0.50 moles of HCl per litre of solution.

Molar mass of Na₂CO₃  

= 2 × 23 + 3 × 16

= 106 g mol -1

0.5 M Na₂CO₃  means 0.05 × 106 g 

= 53g.

26. If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced? 

Ans: H₂ and O₂ react according to the equation:

2H₂(g) + O₂(g) = 2H₂O(g)

The volume of a gas is directly proportional to the number of moles.

Thus, 2 volumes of dihydrogen react with 1 volume of oxygen to form 2 volumes of water.

Therefore, 10 volumes of H₂ will produce 10 volumes of H₂O (water vapour). So, the reaction will produce 10 volumes of water vapour.

27. Convert the following into basic units: 

(i) 28.7 pm.

Ans: 1 picometer (pm) = 10⁻¹² metres

28.7pm=28.7 × 10⁻¹²m / 1pm

=2.87 × 10⁻¹¹m

(ii) 15.15 pm 

Ans: 1 picometer (pm) = 10⁻¹² metres

15.15 us = 15.15 × 10^-12

= 1.515 × 10^-11m

(iii) 25365 m

Ans: 25365 mg = 2.5365 × 1g /1000 mg × 1kg 1000g 

2.5365 × 10-2 kg. Since.

28. Which one of the following will have the largest number of atoms? 

(i) 1 g Au (s)

Ans: 1 g Au (s) =1/197 mol 

= 1/197 × 6.02 × 10²³ atoms

= 6.02 × 10²³ / 197 atoms.

(ii) 1 g Na (s) 

Ans: 1 g Na (s) = 1/23 mol 

= 1/ 23 × 6.02 × 10²³ atoms

= 6.02 × 10²³ / 23 atoms.

(iii) 1 g Li (s) 

Ans: 1 g Li (s) = 1/7 mol 

= 1/ 7 × 6.02 × 10²³ atoms

= 6.02 × 10²³ / 7 atoms.

(iv) 1 g of Cl2(g)

Ans: 1 g of Cl2(g) = 1/71 mol 

= 1/71 × 6.02 × 10²³ atoms

= 6.02 × 10²³ / 71 atoms.

29. 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 component in solution = moles of a particular component /Total number of moles of solution.

 Xc₂H₅OH = n(C₂H₅OH) / n(C₂H₅OH) +  n(H₂O)

0.0040 =  n(C₂H₅OH) / n(C₂H₅OH) +  n(H₂O)

No of moles in 1 L of water

= 1000 g / 18g mol-1

= 55.55 moles

Let n be the number of moles of ethanol. The mole fraction of ethanol is 0.040.

n(C₂H₅OH) / n(C₂H₅OH) + 55.55 = 0.040

Cross-multiply to solve for n:

n = 0.040 (n + 55.55)

n = 0.040n + 2.222

n =  2.222 / 1- 0.040n

n = 2.31 moles.

30. What will be the mass of one ¹²C atom in g? 

Ans: The molar mass ¹²C of is 12 g/mol.

= 12 grams/ 6.022 × 10²³ atoms

= 1.993 × 10^-23 grams.

31. How many significant figures should be present in the answer of the following calculations? 

(i) 0.02856 × 298.15 × 0.112 0.5785 / 0.5785

Ans: The answer of the calculation 0.02856 × 298.15 × 0.112 / 0.5785 the number with the least number of significant figures is 0.1120, which has 3 significant figures.

The least precise number of answer = 0.112

(ii) 5 × 5.364 

Ans: Leaving the exact number (5), the second term has 4 significant figures. Hence, the answer should have 4 significant figures.

(iii) 0.0125 + 0.7864 + 0.0215

Ans: In the given addition, the least number of decimal places in the term is 4. Hence, the answer should have 4 significant.

32. Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes: 

Ans: Molar mass of Ar

= ΣfiAi = 0.00337 × 35.96755 + 0.00063 × 37.96272 + 0.996 × 39.9624 

=  0.1212 + 0.002393 + 39.8025

= 39.926 g mol^-1

∴ Molar mass of naturally occurring argon isotopes is 39.926 g mol^-1

33. Calculate the number of atoms in each of the following: 

(i) 52 moles of Ar.

Ans: 1 mole of Ar = 6.022 × 10²³ atoms of Ar 

= 52 mol of he = 6.022 × 10²³ atoms

= 3.131 × 10²⁵

(ii) 52 u of He.

Ans: Atomic mass of He = 4 amu

∴ 52 u of he = 1/4 × 52 atoms

= 13 atoms.

(iii) 52 g of He. 

Ans: Gram atomic mass of He = 4 g 

Or 4 g of He contains = 6.022 × 10²³ atoms

∴ 52 g of he = 16.022* 10²³ /4 × 52 atoms

= 7.8286 × 10²⁴

34. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate:

(i) Empirical formula.

(ii) Molar mass of the gas. and 

(iii) Molecular formula.

Ans: Moles of Carbon: Given 92.2 g of carbon, moles of carbon = 92.212 = 7.7 mol.

Moles of Hydrogen: Given 7.7 g of hydrogen, moles of hydrogen = 7.71 = 7.7 mol

Mole Ratio: The ratio of C is 7.77.7=1:1.

Calculation Of Empirical Formula: 

Empirical formula = CH 

10.0L of the gas at STP weigh = 11.6

∴ 22.4 L of the gas at S.T.P.  will weight 

= 11.6/ 10.0 × 22.4

= 25.984g

= 26g

∴ Molar mass = 26 g mol^-1 

Empirical formula mass of CH 

= 12+1=13

∴ n = Molecular mass / Empirical formula mass = 26/13

= 2

The molecular formula = 2 × CH = C₂​H₂

35. 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: Calculation of mass of HCl in 25 ml of 0.75 M HCl

1000 mL of 0.75 mol 

= 0.75 × 36.5 g 

= 24.375 g 

∴ 25 mL of 0.75 HCl will contain HCl

= 24.375 g / 1000× 25g 

= 0.6844 g  

CaCO₃ + 2HCl = CaCl₂ + H₂O + CO₂

According to the above equation, 100g of CaCO₃ is required to react completely with CaCO₃. That is

73g of HCl = 100g of CaCO₃

1g of HCl = (100 / 73) g 

= (100 / 73) × 0.684 g 

= 1.369 × 0.684

= 0.937g.

36. 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 MnO₂ i.e

Molar mass of MnO₂ = 55 + 2 × 16

 = 55 + 32 = 87g       

Molar mass of HCl = 1.01 (H) + 35.45 (Cl) = 36.5 g/mol

React with 4 mole of  HCl, i.e., 4 × 36.5 

 = 146 of HCI. 

Therefore, 5.0g of  MnO₂ will react with HCl

= 146/87 × 5.0 

= 8.40 g.