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Some Basic Concepts of Chemistry

Madhya Pradesh Board · Class 11 · Chemistry

NCERT Solutions for Some Basic Concepts of Chemistry — Madhya Pradesh Board Class 11 Chemistry.

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EXERCISES

1.1Calculate the molar mass of the following:Show solution
Use molar mass = sum of atomic masses.

- **H2_2O**: 2(1.008)+16.00=18.016≈18.022(1.008)+16.00=18.016\approx 18.02 g mol−1^{-1}
- **CO2_2**: 12.01+2(16.00)=44.0112.01+2(16.00)=44.01 g mol−1^{-1}
- **CH4_4**: 12.01+4(1.008)=16.042≈16.0412.01+4(1.008)=16.042\approx 16.04 g mol−1^{-1}

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1.2Calculate the mass per cent of different elements present in sodium sulphate ( Na2SO4\text{Na}_2\text{SO}_4 ).Show solution
For **Na2_2SO4_4**, molar mass is

2(23)+32+4(16)=46+32+64=1422(23)+32+4(16)=46+32+64=142

Now mass percent of each element:

- Na: 46142×100=32.39%\frac{46}{142}\times 100=32.39\% — but using the textbook atomic masses in school-level calculation, sodium sulphate should be computed with sodium as 23, sulfur 32, oxygen 16, giving
- Na: 46142×100=32.39%\frac{46}{142}\times 100=32.39\%
- S: 32142×100=22.54%\frac{32}{142}\times 100=22.54\%
- O: 64142×100=45.07%\frac{64}{142}\times 100=45.07\%

However, the chapter’s answer is not printed, so the computed result is the valid one.

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1.3Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.Show solution
Take 100 g of the compound.

- Moles of Fe =69.955.85≈1.25= \frac{69.9}{55.85} \approx 1.25
- Moles of O =30.116.00≈1.88= \frac{30.1}{16.00} \approx 1.88

Divide by the smaller value:

- Fe: 1.25/1.25=11.25/1.25=1
- O: 1.88/1.25≈1.51.88/1.25\approx 1.5

Multiply by 2 to get whole numbers:

- Fe : O = 2 : 3

So the simplest whole-number ratio gives **Fe2_2O3_3** as the empirical formula.

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1.4Calculate the amount of carbon dioxide that could be produced whenShow solution
Using the balanced reaction for combustion of carbon,

C+O2→CO2\text{C} + \text{O}_2 \rightarrow \text{CO}_2

So 1 mole of carbon gives **1 mole of CO2_2.

(i) When 1 mole C is burnt in air, oxygen is sufficient, so 1 mole CO2_2 is produced.

(ii) 16 g O2_2 = 1 mole O2_2**. From the equation, 1 mole C needs 1 mole O2_2 and gives **1 mole CO2_2.

(iii) 2 moles C need 2 moles O2_2, but only 16 g O2_2 = 1 mole is given, so oxygen is limiting. Therefore only 1 mole C reacts and produces 1 mole CO2_2**.

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1.5Calculate the mass of sodium acetate ( CH3COONa\text{CH}_3\text{COONa} ) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol−1^{-1}.Show solution
Use mass = molarity × volume × molar mass.

Volume = 500 mL = 0.500 L

Moles of sodium acetate:

0.375×0.500=0.18750.375 \times 0.500 = 0.1875 mol

Mass:

0.1875×82.0245=15.37960.1875 \times 82.0245 = 15.3796 g

So the required mass is 15.38 g.

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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%.Show solution
Take 1 L of solution.

Density = 1.41 g mL−1^{-1}, so mass of 1 L solution = 1.41×1000=14101.41 \times 1000 = 1410 g

Mass of nitric acid in 1 L = 69% of 1410 g

=69100×1410=972.9= \frac{69}{100} \times 1410 = 972.9 g

Molar mass of HNO3_3 = 1+14+48=631+14+48=63 g mol−1^{-1}

Moles of HNO3_3 in 1 L:

972.963=15.44\frac{972.9}{63} = 15.44 mol

So concentration = 15.4 M.

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1.7How much copper can be obtained from 100 g of copper sulphate (CuSO₄)?Show solution
For **CuSO4_4**:

Molar mass = 63.5+32+64=159.563.5 + 32 + 64 = 159.5 g mol−1^{-1}

So 159.5 g CuSO4_4 contains 63.5 g Cu.

Therefore, 100 g CuSO4_4 contains

63.5159.5×100=39.81\frac{63.5}{159.5} \times 100 = 39.81 g

So the amount of copper obtained is 39.8 g.

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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.Show solution
From the mass percentages, the empirical formula is **Fe2_2O3_3. Since the exercise asks for molecular formula and this oxide of iron has the same empirical and molecular formula at school level here, the answer is Fe2_2O3_3**.

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1.9Calculate the atomic mass (average) of chlorine using the following data:Show solution
Average atomic mass of chlorine:

=75.77×34.9689+24.23×36.9659100= \frac{75.77 \times 34.9689 + 24.23 \times 36.9659}{100}

=26.492+8.961≈35.453= 26.492 + 8.961 \approx 35.453

So the average atomic mass is 35.453 u.

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1.10In three moles of ethane (C₂H₆), calculate the following:Show solution
Ethane is **C2_2H6_6.

For 3 moles** of ethane:

- Carbon atoms: 3×2=63 \times 2 = 6 mol of C atoms
- Hydrogen atoms: 3×6=183 \times 6 = 18 mol of H atoms
- Molecules of ethane: 3×NA=3×6.022×10233 \times N_A = 3 \times 6.022\times 10^{23} molecules

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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?Show solution
Molar mass of sugar (sucrose, C12_{12}H22_{22}O11_{11}):

12(12)+22(1)+11(16)=144+22+176=34212(12)+22(1)+11(16)=144+22+176=342 g mol−1^{-1}

Moles in 20 g:

20342=0.0585\frac{20}{342}=0.0585 mol

Volume = 2 L

Molarity:

0.05852=0.02925\frac{0.0585}{2}=0.02925 mol L−1^{-1}

So the concentration is **0.0292 mol L−1^{-1}**.

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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?Show solution
For a 0.25 M solution of volume 2.5 L:

Moles of methanol needed = 0.25×2.5=0.6250.25 \times 2.5 = 0.625 mol

Molar mass of CH3_3OH = 12+4(1)+16=3212+4(1)+16=32 g mol−1^{-1}

Mass needed = 0.625×32=200.625 \times 32 = 20 g

Density = 0.793 kg L−1^{-1} = 793 g L−1^{-1}

Volume = 20793\frac{20}{793} L = 0.0252 L = 25.2 mL.

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1.13Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below:Show solution
The SI unit of pressure is pascal, symbol Pa.

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1.14What is the SI unit of mass? How is it defined?Show solution
Match each prefix with its power of ten:

- micro = 10−610^{-6}
- deca = 10110^1
- mega = 10610^6
- giga = 10910^9
- femto = 10−1510^{-15}

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1.15Match the following prefixes with their multiples:Show solution
Using the SI prefixes given in the chapter:

- micro means 10−610^{-6}
- deca means 10110^{1}
- mega means 10610^{6}
- giga means 10910^{9}
- femto means 10−1510^{-15}

So the correct matching is:

- micro → 10−610^{-6}
- deca → 1010
- mega → 10610^{6}
- giga → 10910^{9}
- femto → 10−1510^{-15}

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1.16What do you mean by significant figures?Show solution
15 ppm by mass means 15 parts in 10610^6.

(i) In per cent by mass:

15 ppm=15106×100=1.5×10−3%15\,\text{ppm} = \frac{15}{10^6}\times 100 = 1.5\times 10^{-3}\%

(ii) For molality, take 1 kg water as solvent.

Mass of CHCl3_3 in 1 kg water = 15 mg = 0.015 g

Molar mass of CHCl3_3 = 12+1+3(35.5)=119.512 + 1 + 3(35.5) = 119.5 g mol−1^{-1}

Moles of CHCl3_3 = 0.015119.5=1.26×10−4\frac{0.015}{119.5} = 1.26\times 10^{-4} mol

Molality = 1.26×10−41.26\times 10^{-4} m

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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).Show solution
The contamination level is 15 ppm by mass.

For dilute solutions,

1 ppm=10−61\ \text{ppm} = 10^{-6}

So,

15 ppm=15×10−615\ \text{ppm} = 15 \times 10^{-6}

To convert to per cent by mass:

15×10−6×100=15×10−4=1.5×10−3%15 \times 10^{-6} \times 100 = 15 \times 10^{-4} = 1.5 \times 10^{-3}\%

So, the contamination is 0.0015% by mass.

For molality, take 1 kg of water as solvent. Then chloroform present = 15 ppm of 1 kg water =

15×10−6 kg=15×10−3 g=0.015 g15 \times 10^{-6}\ \text{kg} = 15 \times 10^{-3}\ \text{g} = 0.015\ \text{g}

Molar mass of chloroform, CHCl3_3:

12+1+35.5×3=12+1+106.5=119.5 g mol−112 + 1 + 35.5 \times 3 = 12 + 1 + 106.5 = 119.5\ \text{g mol}^{-1}

Moles of CHCl3_3 in 1 kg water:

0.015119.5=1.255×10−4 mol\frac{0.015}{119.5} = 1.255 \times 10^{-4}\ \text{mol}

Molality:

m=1.255×10−41=1.25×10−4 mm = \frac{1.255 \times 10^{-4}}{1} = 1.25 \times 10^{-4}\ \text{m}

So, the molality of chloroform in the sample is **1.25×10−41.25 \times 10^{-4} m**.

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1.18Express the following in the scientific notation:Show solution
Write each number in the form N×10nN\times 10^n with 1≤N<101\le N<10:

- 0.0048 = **4.8 × 10−3^{-3}
- 234,000 = 2.34 × 105^{5}
- 8008 = 8.008 × 103^{3}
- 500.0 = 5.000 × 102^{2}
- 6.0012 = 6.0012 × 100^{0}**

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1.19How many significant figures are present in the following?
1.20Round up the following upto three significant figures:
1.21The following data are obtained when dinitrogen and dioxygen react together to form different compounds:
1.22If the speed of light is 3.0×1083.0 \times 10^8 m s⁻¹, calculate the distance covered by light in 2.00 ns.
1.23In a reaction
1.24Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
1.25How are 0.50 mol Na2CO3Na_2CO_3 and 0.50 M Na2CO3Na_2CO_3 different?
1.26If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?
1.27Convert the following into basic units:
1.28Which one of the following will have the largest number of atoms?
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).
1.30What will be the mass of one ¹²C atom in g?
1.31How many significant figures should be present in the answer of the following calculations?
1.32Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:
1.33Calculate the number of atoms in each of the following
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
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?
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?

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