CBSE · NCERT · Class 11 Chemistry · Chapter 1

NCERT Solutions: Class 11 Chemistry Chapter 1 - Some Basic Concepts of Chemistry

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Chapter-wise NCERT intext questions and exercise answers for Some Basic Concepts of Chemistry, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
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Exercises 32
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1Exercises32 questions
Q.1.1Calculate the molar mass of the following: (i) H2O (ii) CO2 (iii) CH4v
Solution

Use approximate atomic masses H = 1.008, C = 12.01 and O = 16.00. H2O: 2(1.008)+16.00 = 18.02 g mol^-1. CO2: 12.01+2(16.00)=44.01 g mol^-1. CH4: 12.01+4(1.008)=16.04 g mol^-1.

Answer:

(i) H2O = 18.02 g mol^-1, (ii) CO2 = 44.01 g mol^-1, (iii) CH4 = 16.04 g mol^-1.

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Q.1.2Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4).v
Solution

Molar mass of Na2SO4 = 2(22.99)+32.06+4(16.00)=142.04 g mol^-1. Mass percent Na = 45.98/142.04 x 100 = 32.37%. Mass percent S = 32.06/142.04 x 100 = 22.57%. Mass percent O = 64.00/142.04 x 100 = 45.06%.

Answer:

Na = 32.37%, S = 22.57%, O = 45.06%.

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Q.1.3Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.v
Solution

In 100 g compound, moles of Fe = 69.9/55.85 = 1.252 and moles of O = 30.1/16.00 = 1.881. Dividing by 1.252 gives Fe:O = 1:1.50. Multiplying by 2 gives 2:3, so the empirical formula is Fe2O3.

Answer:

Fe2O3.

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Q.1.4Calculate 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.v
Solution

The reaction is C + O2 -> CO2. (i) 1 mol C gives 1 mol CO2 = 44 g. (ii) 16 g O2 = 0.5 mol O2, so oxygen is limiting and gives 0.5 mol CO2 = 22 g. (iii) With 2 mol C and only 0.5 mol O2, oxygen is again limiting and gives 0.5 mol CO2 = 22 g.

Answer:

(i) 44 g CO2. (ii) 22 g CO2. (iii) 22 g CO2.

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Q.1.5Calculate 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.v
Solution

Moles required = M x V = 0.375 mol L^-1 x 0.500 L = 0.1875 mol. Mass = 0.1875 x 82.0245 = 15.38 g.

Answer:

15.38 g.

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Q.1.6Calculate 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%.v
Solution

Mass of 1 L solution = 1.41 g mL^-1 x 1000 mL = 1410 g. Mass of HNO3 = 69% of 1410 = 972.9 g. Molar mass of HNO3 = 63.01 g mol^-1, so moles = 972.9/63.01 = 15.4 mol. Therefore concentration = 15.4 mol L^-1.

Answer:

15.4 mol L^-1.

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Q.1.7How much copper can be obtained from 100 g of copper sulphate (CuSO4)?v
Solution

Molar mass of CuSO4 = 63.55+32.06+4(16.00)=159.61 g mol^-1. Copper fraction = 63.55/159.61. From 100 g CuSO4, copper obtained = 100 x 63.55/159.61 = 39.8 g.

Answer:

39.8 g copper.

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Q.1.8Determine 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.v
Solution

For 100 g oxide, moles Fe = 69.9/55.85 = 1.252 and moles O = 30.1/16.00 = 1.881. The simplest ratio is Fe:O = 1:1.5 = 2:3. Therefore the formula of the oxide is Fe2O3.

Answer:

Fe2O3.

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Q.1.9Calculate the atomic mass (average) of chlorine using the following data: % Natural Abundance Molar Mass 35Cl 75.77 34.9689 37Cl 24.23 36.9659v
Solution

Average atomic mass = (75.77 x 34.9689 + 24.23 x 36.9659)/100 = 35.45 u.

Answer:

35.45 u.

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Q.1.10In three moles of ethane (C2H6), calculate the following: (i) Number of moles of carbon atoms. (ii) Number of moles of hydrogen atoms. (iii) Number of molecules of ethane.v
Solution

Each mole of C2H6 contains 2 mol C atoms and 6 mol H atoms. Therefore 3 mol C2H6 contains 6 mol C atoms and 18 mol H atoms. Number of ethane molecules = 3 x 6.022 x 10^23 = 1.81 x 10^24.

Answer:

(i) 6 mol C atoms. (ii) 18 mol H atoms. (iii) 1.81 x 10^24 molecules of ethane.

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Q.1.11What 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?v
Solution

Molar mass of C12H22O11 = 12(12.01)+22(1.008)+11(16.00) = about 342 g mol^-1. Moles = 20/342 = 0.0585 mol. Molarity = 0.0585/2.0 = 0.029 mol L^-1.

Answer:

0.029 mol L^-1.

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Q.1.12If 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?v
Solution

Moles of methanol needed = 0.25 x 2.5 = 0.625 mol. Molar mass of CH3OH is about 32.04 g mol^-1, so mass = 0.625 x 32.04 = 20.0 g = 0.0200 kg. Volume = mass/density = 0.0200/0.793 = 0.0252 L = 25.2 mL.

Answer:

0.0252 L, or 25.2 mL.

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Q.1.13Pressure 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.v
Solution

1034 g cm^-2 = 1.034 kg per 10^-4 m^2 = 1.034 x 10^4 kg m^-2. Pressure = weight/area = (1.034 x 10^4)(9.8) = 1.01 x 10^5 N m^-2 = 1.01 x 10^5 Pa.

Answer:

1.01 x 10^5 Pa.

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Q.1.15Match the following prefixes with their multiples: Prefixes Multiples (i) micro 106 (ii) deca 109 (iii) mega 10–6 (iv) giga 10–15 (v) femto 10v
Solution

SI prefixes denote powers of ten: micro means 10^-6, deca means 10, mega means 10^6, giga means 10^9 and femto means 10^-15.

Answer:

(i) micro = 10^-6; (ii) deca = 10; (iii) mega = 10^6; (iv) giga = 10^9; (v) femto = 10^-15.

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Q.1.16What do you mean by significant figures?v
Solution

Measurements contain uncertainty. Significant figures show the precision of a measurement by retaining the digits known with certainty plus one estimated digit.

Answer:

Significant figures are the meaningful digits in a measured or calculated value, including all certain digits and the first uncertain digit.

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Q.1.17A 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.v
Solution

15 ppm by mass means 15 parts in 10^6 parts by mass. Percent by mass = 15/10^6 x 100 = 0.0015%. For 1 kg water sample, mass of CHCl3 is 15 mg = 0.015 g. Molar mass of CHCl3 = 12.01+1.008+3(35.45)=119.37 g mol^-1, so moles = 0.015/119.37 = 1.26 x 10^-4 mol. The solvent mass is essentially 1 kg, so molality = 1.26 x 10^-4 mol kg^-1.

Answer:

(i) 0.0015% by mass. (ii) 1.26 x 10^-4 mol kg^-1.

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Q.1.18Express the following in the scientific notation: (i) 0.0048 (ii) 234,000 (iii) 8008 (iv) 500.0 (v) 6.0012v
Solution

Scientific notation writes a number as a x 10^n, where 1 <= a < 10. Moving the decimal point gives the listed powers of ten.

Answer:

(i) 4.8 x 10^-3; (ii) 2.34 x 10^5; (iii) 8.008 x 10^3; (iv) 5.000 x 10^2; (v) 6.0012 x 10^0.

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Q.1.19How many significant figures are present in the following? (i) 0.0025 (ii) 208 (iii) 5005 (iv) 126,000 (v) 500.0 (vi) 2.0034v
Solution

Leading zeros are not significant; zeros between non-zero digits are significant; zeros after a decimal point are significant. For 126,000 without a decimal point, the trailing zeros are not taken as significant, so it has 3 significant figures.

Answer:

(i) 2; (ii) 3; (iii) 4; (iv) 3; (v) 4; (vi) 5.

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Q.1.20Round up the following upto three significant figures: (i) 34.216 (ii) 10.4107 (iii) 0.04597 (iv) 2808v
Solution

Keep three significant figures and round using the next digit. 34.216 -> 34.2; 10.4107 -> 10.4; 0.04597 -> 0.0460, where the final zero is significant; 2808 -> 2810, better written as 2.81 x 10^3 to show three significant figures.

Answer:

(i) 34.2; (ii) 10.4; (iii) 0.0460; (iv) 2.81 x 10^3.

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Q.1.21The following data are obtained when dinitrogen and dioxygen react together to form different compounds: Mass of dinitrogen Mass of dioxygen (i) 14 g 16 g (ii) 14 g 32 g (iii) 28 g 32 g (iv) 28 g 80 g (a) Which law of chemical combination is obeyed by the above experimental data? Give its statement. (b) Fill in the blanks in the following conversions: (i) 1 km = ...................... mm = ...................... pm (ii) 1 mg = ...................... kg = ...................... ng (iii) 1 mL = ...................... L = ...................... dm3v
Solution

For a fixed mass of dinitrogen, the masses of dioxygen combining with it are in small whole-number ratios. Converting all data to 14 g dinitrogen gives oxygen masses 16, 32, 16 and 40 g, which are in the ratio 2:4:2:5 after division by 8. This illustrates the law of multiple proportions: when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in simple whole-number ratios. Unit conversions follow from kilo = 10^3, milli = 10^-3, pico = 10^-12, nano = 10^-9 and 1 L = 1 dm^3.

Answer:

(a) Law of multiple proportions. (b) (i) 1 km = 10^6 mm = 10^15 pm; (ii) 1 mg = 10^-6 kg = 10^6 ng; (iii) 1 mL = 10^-3 L = 10^-3 dm^3.

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Q.1.22If the speed of light is 3.0 × 108 m s–1, calculate the distance covered by light in 2.00 ns.v
Solution

2.00 ns = 2.00 x 10^-9 s. Distance = speed x time = (3.0 x 10^8)(2.00 x 10^-9) = 0.600 m.

Answer:

0.600 m.

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Q.1.25How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?v
Solution

Mole measures quantity, independent of volume. Molarity measures moles of solute per litre of solution, so it specifies concentration.

Answer:

0.50 mol Na2CO3 is an amount of substance; 0.50 M Na2CO3 is a concentration equal to 0.50 mol Na2CO3 per litre of solution.

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Q.1.26If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?v
Solution

For gases at the same temperature and pressure, volume ratios follow mole ratios. The balanced equation is 2H2 + O2 -> 2H2O. Therefore 10 volumes H2 react with 5 volumes O2 to produce 10 volumes H2O vapour.

Answer:

10 volumes of water vapour.

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Q.1.27Convert the following into basic units: (i) 28.7 pm (ii) 15.15 pm (iii) 25365 mgv
Solution

1 pm = 10^-12 m and 1 mg = 10^-6 kg. Therefore 28.7 pm = 28.7 x 10^-12 m = 2.87 x 10^-11 m; 15.15 pm = 1.515 x 10^-11 m; 25365 mg = 25365 x 10^-6 kg = 2.5365 x 10^-2 kg.

Answer:

(i) 2.87 x 10^-11 m. (ii) 1.515 x 10^-11 m. (iii) 2.5365 x 10^-2 kg.

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Q.1.28Which one of the following will have the largest number of atoms? (i) 1 g Au (s) (ii) 1 g Na (s) (iii) 1 g Li (s) (iv) 1 g of Cl2(g)v
  1. i. 1 g Au (s)
  2. ii. 1 g Na (s)
  3. iii. 1 g Li (s)
  4. iv. 1 g of Cl2(g)
Solution

For elements, atoms in 1 g are proportional to 1/atomic mass. Lithium has the smallest atomic mass among Au, Na and Li, so 1 g Li contains the most atoms. For Cl2, 1 g gives 2/70.9 mol atoms, still less than 1/6.94 mol atoms for Li.

Answer:

(iii) 1 g Li (s).

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Q.1.29Calculate 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).v
Solution

Assume 1 L water, so mass of water = 1000 g and moles of water = 1000/18 = 55.56 mol. If mole fraction of ethanol is 0.040, then n_ethanol/(n_ethanol+55.56)=0.040. Thus n_ethanol = (0.040/0.960)(55.56)=2.31 mol. Taking the solution volume approximately as 1 L, molarity = 2.31 M.

Answer:

2.31 M approximately.

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Q.1.30What will be the mass of one 12C atom in g?v
Solution

One mole of 12C atoms has mass 12 g and contains 6.022 x 10^23 atoms. Mass of one atom = 12/(6.022 x 10^23) = 1.99 x 10^-23 g.

Answer:

1.99 x 10^-23 g.

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Q.1.32Use the data given in the following table to calculate the molar mass of naturally occuring argon isotopes: Isotope Isotopic molar mass Abundance 36Ar 35.96755 g mol–1 0.337% 38Ar 37.96272 g mol–1 0.063% 40Ar 39.9624 g mol–1 99.600%v
Solution

Average molar mass = 35.96755(0.00337)+37.96272(0.00063)+39.9624(0.99600) = 39.95 g mol^-1.

Answer:

39.95 g mol^-1.

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Q.1.33Calculate the number of atoms in each of the following (i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He.v
Solution

(i) 52 mol Ar contains 52 x 6.022 x 10^23 = 3.13 x 10^25 atoms. (ii) One He atom has mass about 4 u, so 52 u He contains 52/4 = 13 atoms. (iii) 52 g He = 52/4 = 13 mol He, so atoms = 13 x 6.022 x 10^23 = 7.83 x 10^24.

Answer:

(i) 3.13 x 10^25 atoms. (ii) 13 atoms. (iii) 7.83 x 10^24 atoms.

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Q.1.34A 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.v
Solution

Moles C from CO2 = 3.38/44.01 = 0.0768 mol. Moles H from water = 2(0.690/18.015)=0.0766 mol. Ratio C:H is approximately 1:1, so empirical formula is CH and empirical formula mass is about 13 g mol^-1. At STP, 10.0 L gas = 10.0/22.4 = 0.446 mol. Molar mass = 11.6/0.446 = 26.0 g mol^-1. Molecular formula factor = 26/13 = 2, so molecular formula is C2H2.

Answer:

(i) CH. (ii) 26.0 g mol^-1. (iii) C2H2.

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Q.1.35Calcium 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?v
Solution

Moles HCl = 0.75 x 0.025 = 0.01875 mol. From CaCO3 + 2HCl -> CaCl2 + CO2 + H2O, moles CaCO3 required = 0.01875/2 = 0.009375 mol. Molar mass CaCO3 is about 100.09 g mol^-1, so mass = 0.009375 x 100.09 = 0.94 g.

Answer:

0.94 g CaCO3.

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Q.1.36Chlorine 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?v
Solution

Molar mass of MnO2 = 54.94+2(16.00)=86.94 g mol^-1. Moles MnO2 = 5.0/86.94 = 0.0575 mol. The reaction requires 4 mol HCl per mol MnO2, so moles HCl = 4(0.0575)=0.230 mol. Mass HCl = 0.230 x 36.46 = 8.4 g.

Answer:

8.4 g HCl.

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