Chapter 1
Some Basic Concepts of Chemistry
1. Calculate the molecular mass of the following :
(i) H2O
(ii) CO2
(iii) CH4
Solution :
(i)CH4 :Molecular weight of methane, CH4= (1 x Atomic weight of carbon) + (4 x Atomic weight of hydrogen)= [1(12.011 u) +4 (1.008u)]= 12.011u + 4.032 u= 16.043 u
(ii) H2O :Molecular weight of water, H2O= (2 x Atomic weight of hydrogen) + (1 x Atomic weight of oxygen)= [2(1.0084) + 1(16.00 u)]= 2.016 u +16.00 u= 18.016uSo approximately= 18.02 u
(iii) CO2 := Molecular weight of carbon dioxide, CO2= (1 x Atomic weight of carbon) + (2 x Atomic weight of oxygen)= [1(12.011 u) + 2(16.00 u)]= 12.011 u +32.00 u= 44.011 uSo approximately= 44.01u
2. Calculate the mass percent of different elements present in Sodium Sulphate (Na2SO4).
Solution :
Molar mass of Na2SO4 = [(2 × 23.0) + (32.00) + 4 (16.00)] = 142 gMass percent of an element = (Mass of that element in compound/Molar mass of that compound) × 100∴ Mass percent of sodium (Na): (46/142) × 100 = 32.39%Mass percent of sulphur(S): (32/142) × 100 = 22.54%Mass percent of oxygen:(O): (64/142) × 100 = 45.07%
3. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
Solution :
% of iron by mass = 69.9 % [Given]% of oxygen by mass = 30.1 % [Given]Atomic mass of iron = 55.85 amuAtomic mass of oxygen = 16.00 amuRelative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25Relative moles of oxygen in iron oxide = %mass of oxygen by mass/Atomic mass of oxygen = 30.01/16=1.88Simplest molar ratio = 1.25/1.25 : 1.88/1.25 ⇒ 1 : 1.5 = 2 : 3
∴ The empirical formula of the iron oxide is 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.
Solution :
The balanced reaction of combustion of carbon in dioxygens is:C(s) + O2(g) → CO2 (g)1mole 1mole(32g) 1mole(44g)(i) In dioxygen, combustion is complete. Therefore 1 mole of carbon dioxide produced by burning 1 mole of carbon.
(ii) Here, oxgen acts as a limiting reagent as only 16 g of dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide.
(iii) Here again oxgen acts as a limiting reagent as only 16 g of dioxygen is available. It is a limiting reactant. Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.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 82.0245g mol-1.Solution :0.375 M aqueous solution of sodium acetate means that 1000 mL of solution containing 0.375 moles of sodium acetate.∴No. of moles of sodium acetate in 500 mL = (0.375/1000)×500 = 0.375/2 = 0.1875Molar mass of sodium acetate = 82.0245g mol-1∴Mass of sodium acetate acquired = 0.1875×82.0245 g = 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%.
Solution :
Mass percent of 69% means tat 100g of nitric acid solution contain 69 g of nitric acid by mass.Molar mass of nitric acid (HNO3) = 1+14+48 = 63g mol-1Number of moles in 69 g of HNO3 = 69/63 moles = 1.095 molesVolume of 100g nitric acid solution = 100/1.41 mL = 70.92 mL = 0.07092 L∴ Conc. of HNO3 in moles per litre = 1.095/0.07092 = 15.44 M
7. How much copper can be obtained from 100 g of copper sulphate (CuSO4 )?
Solution :
1 mole of CuSO4 contains 1 mole of copper.Molar mass of CuSO4 = (63.5) + (32.00) + 4(16.00)= 63.5 + 32.00 + 64.00 = 159.5 g159.5 g of CuSO4 contains 63.5 g of copper.∴ copper can be obtained from 100 g of copper sulphate = (63.5/159.5)×100 = 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.Given that the molar mass of the oxide is 159.69 g mol- 1
Solution :
% of iron by mass = 69.9 % [Given]% of oxygen by mass = 30.1 % [Given]Atomic mass of iron = 55.85 amu Atomic mass of oxygen = 16.00 amu Relative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25Relative moles of oxygen in iron oxide = %mass of oxygen by mass /Atomic mass of oxygen = 30.01/16 =1.88Simplest molar ratio = 1.25/1.25 : 1.88/1.25 ⇒ 1 : 1.5 = 2 : 3∴ The empirical formula of the iron oxide is Fe2O3.Mass of Fe 2O3 = (2×55.85) + (3×16.00) = 159.7 g mol- 1n = Molar mass /Empirical formula mass = 159.7/ 159.6 = 1(approx)Thus, Molecular formula is same as Empirical Formula i.e. Fe2O3.
9. Calculate the atomic mass (average) of chlorine using the following data : % Natural Abundance Molar Mass35Cl 75.77 34.968937Cl 24.23 36.9659Solution :Fractional Abundance of 35Cl = 0.7577 and Molar mass = 34.9689Fractional Abundance of 37Cl = 0.2423 and Molar mass = 36.9659∴ Average Atomic mass = (0.7577 × 34.9689) amu + (0.2423 × 36.9659)= 26.4959 + 8.9568 = 35.4527
10. In 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.Solution :(i) 1 mole of C2H6 contains 2 moles of Carbon atoms∴ 3 moles of of C2H6 will contain 6 moles of Carbon atoms
(ii) 1 mole of C2H6 contains 6 moles of Hydrogen atoms∴ 3 moles of of C2H6 will contain 18 moles of Hydrogen atoms
(iii) 1 mole of C2H6 contains Avogadro’s no. 6.02 ×1023 molecules∴ 3 moles of of C2H6 will contain ethane molecule = 3×6.02 × 1023= 18.06 ×1023 molecules.
11. What is the concentration of sugar (C12H22O11) in mol L-1 if its 20 g are dissolved inenough water to make a final volume up to 2L?
Solution :
Molar mass of sugar (C12H22O11) = (12 ×12) +(1 ×22)+ (11×16) = 342 g mol-1No. of moles in 20g of sugar = 20/342 = 0.0585 moleVolume of Solution = 2L (given)Molar concentration = Moles of solute/Volume of solution in L = 0.0585mol /2L = 0.0293 mol L-1 = 0.0293 M
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?
Solution :
Molar mass of methanol (CH3OH) = (1×12) + (4×1) + (1×16) = 32 g mol-1 = 0.032 kg mol-1Molarity of the solution = 0.793 /0.032 = 24.78 mol L-1 Applying, M1V1 (Given Solution) = M2V2 (Solution to be prepared)24.78×V1 = 0.25×2.5 LV1= 0.02522 L = 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-2If mass of air at sea level is 1034 g cm-2,calculate the pressure in pascal.
Solution :
Pressure is the force (i.e. weigh) acting per unit area.P= F/A = 1034g × 9.8ms -2/cm2 = 1034g × 9.8ms-2 /cm2 × 1kg /1000g × 100cm /1m × 100cm /1m = 1.01332 ×105 NNow,1Pa = 1N m-2∴ 1.01332 × 105 N ×m-2 = 1.01332 ×105 Pa
14. What is the SI unit of mass? How is it defined?
Solution :
The SI unit of mass is kilogram (kg).The kg is defined as the mass of platinum-iridium (Pt-Ir) cylinder that is stored in an air-tight jar at International Bureau of Weigh and Measures in France.
15. Match the following prefixes with their multiples:
Prefixes Multiples
(a)femto10
(b)giga10−15
(c)mega10−6
(d)deca109
(e)micro106
Solution :
PrefixesMultiples
(a)femto10−15
(b)giga109
(c)mega106
(d)deca10
(e)micro10−6
16. What do you mean by significant figures ?
Solution :
Significant figures are meaningful digits which are known with certainty including the last digit whose value is uncertain.
For example,
In 11.2546 g, there are 6 significant figures but here 11.254 is certain and 6 is uncertain and the uncertainty would be ±1 in the last digit. Hence last uncertain digit is also included in Significant figures.
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 percent by mass.
(ii) Determine the molality of chloroform in the water sample.
Solution :
(i) 15 ppm means 5 parts in million(106) parts.∴ % by mass = 15/106 × 100 = 15 × 10-4 = 1.5×10-3 %
(ii) Molar mass of chloroform(CHCl3) = 12+1+ (3×35.5) = 118.5 g mol-1100g of the sample contain chloroform = 1.5×10-3g∴ 1000 g(1 kg) of the sample will contain chloroform = 1.5×10-2 g= 1.5×10-2/ 118.65 mole = 1.266 ×10-4 mole∴ Molality = 1.266×10-4 m.
18. Express the following in the scientific notation:(i) 0.0048(ii) 234,000(iii) 8008(iv) 500.0(v) 6.0012
Solution :
(i) 0.0048 = 4.8× 10-3
(ii) 234, 000 = 2.34× 105
(iii) 8008 = 8.008× 103
(iv) 500.0 = 5.000× 102
(v) 6.0012 = 6.0012× 10019.
How many significant figures are present in the following?
(i) 0.0025
(ii) 208
(iii) 5005
(iv) 126,000
(v) 500.0
(vi) 2.0034
Solution :
(i) 2
(ii) 3
(iii) 4
(iv) 3
(v) 4
(vi) 520.
Round up the following upto three significant figures:
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808
Solution :
(i) 34.2
(ii) 10.4
(iii) 0.046
(iv) 2810