This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Explain : Molar mass. |
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Answer» i. The mass of one mole of a substance (element/compound) in grams is called its molar mass. ii. The molar mass of any element in grams is numerically equal to atomic mass of that element in u. e.g.
iii. Similarly, molar mass of polyatomic molecule, in grams is numerically equal to its molecular mass or formula mass in u. e.g.
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| 2. |
Explain :The need of the term average atomic mass. |
Answer»
Hence, The term average atomic mass is needed to express atomic mass of elements containing mixture of two or more isotopes. |
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| 3. |
What do you mean by Avagadro Number? |
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Answer» The number of molecules present in 1 gm molecular weight of any substance. OR The number of atoms present in 1 gm atomic weight of any element. It is represented by the symbol N – 6.022 × 1023. |
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| 4. |
What is meant by molar volume of a gas? |
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Answer» The volume occupied by one mole of a gas at standard temperature (0°C) and pressure (1 atm) (STP) is called as molar volume of a gas. The molar volume of a gas at STP is 22.4 dm3. |
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| 5. |
What do you understand by the terms (i) empirical formula and (ii) molecular formula? How are they related o each other? Illustrate with an example. |
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Answer» (i) Empirical formula: The formula which gives the simple whole number ratio of the atoms of various elements present in one molecule of the compound is called empirical formula. For example, empirical formula of hydrogen peroxide is HO. It represents that hydrogen and oxygen (i.e., H : O) are present in the ratio of 1 : 1 in hydrogen peroxide. (ii) Molecular formula: The formula which gives the actual number of atoms of various elements present in one molecule of the compound is called molecular formula. For example, the molecular formula of hydrogen. peroxide is H2O2 because one molecule of hydrogen peroxide contains two atoms of hydrogen and two atoms of oxygen. Relation between Empirical and Molecular Formulae: Molecular formula = n (Empirical formula) where n is a simple whole number and may have values 1, 2, 3.... It is equal to n = {Molecular mass}/{Empirical formula mass} |
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| 6. |
Explain limiting and excess reagents. |
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Answer» In many situations, an excess of one or more substance is available for chemical reaction. Some of these excess substances will therefore, be left over when the reaction is complete. The reaction stops immediately as soon as one of the reactants is totally consumed. Consider a chemical reaction given below initiated by passing a spark through a reaction vessel containing 10 mol of H2 and 7 mol of O2. Since 2 mol H2 = 1 mol O2, thus 2H2 (g) + O2 (g) → 2H2O (g) Moles before reaction 10 7 0 Moles after reaction 0 2 10 It is thus evident that the reaction stops only after consumption of 5 moles O2 since no further amount of H2 is left to react with unreacted O2. The substance (here H2) that is completely consumed in the reaction is called limiting reagent because it determines or limits, the amount of product. The other reactants present in excess are called as excess reagents. |
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| 7. |
Calculate the molecular mass of the following in u. a. NH3 b. CH3COOH c. C2H5OH |
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Answer» (a) Molecular mass of NH3 = (1 × Average atomic mass of N) + (3 × Average atomic mass of H) = (1 × 14.0 u) +(3 × 1.0 u) = 17 u (b) Molecular mass of CH3COOH = (2 × Average atomic mass of C) + (4 × Average atomic mass of H) + (2 × Average atomic mass of O) = (2 × 12.0 u) + (4 × 1.0 u) + (2 × 16.0 u) = 60 u (c). Molecular mass of C2H5OH = (2 × Average atomic mass of C) + (6 × Average atomic mass of H) + (1 × Average atomic mass of O) = (2 × 12.0 u) + (6 × 1.0 u) + (1 × 16.0 u) = 46 u |
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| 8. |
Express the result of the following calculations to the appropriate number of significant figures(i) {3.24 x 0.08666}/{5.006}(ii) 0.58 + 324.65(iii) 1.78986 x 103(iv) 943 x 0.00345 + 101 |
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Answer» (i) {3.24 x 0.08666}/{5.006} = {0.2807784}/{5.006} = 0.0561 (ii) 0.58 + 324.65 = 325.23 (iii) 1.78986 x 103 = 1789.86 (iv) 943 x 0.00345 + 101 = 3.25 + 101 = 104.25 |
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| 9. |
What is the SI unit of amount of a substance? |
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Answer» The SI unit for the amount of a substance is mole (mol). |
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| 10. |
What is a balanced chemical equation. |
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Answer» A chemical equation which has an equal number of atoms of each element in the reactants and the products is called balanced chemical equation. |
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| 11. |
Explain how mixture differ from pure substances? |
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| 12. |
What is chemistry? |
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Answer» Chemistry is the branch of science that studies the composition, properties and interaction of matter. |
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| 13. |
How is mole related to (i) mass (ii) volume and (iii) number of molecules of a substance? |
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Answer» (i) One mole of atoms = Gram atomic mass of the element One mole of molecules = gram molecular mass Mass of one atom = {Atomic mass}/{6.023 x 1023} (ii) Volume occupied by 1 mole of a gas at N.T.P. = 22.4 L (iii) One mole of molecules = 6.023 x 1023 molecules. |
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| 14. |
How many particles are present in 1 mole of a substance? |
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Answer» The number of particles in one mole is 6.0221367 × 1023. |
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| 15. |
What is molar mass? |
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Answer» The mass of one mole of a substance is called its molar mass. |
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| 16. |
What is the basic unit of mass in the SI system? |
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Answer» The basic unit of mass in the SI system is kilogram (kg). |
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| 17. |
Name the following : i. Full form of CGS unit system ii. Full form of FPS unit system iii. The SI unit of length iv. Symbol used for Candela unit v. SI unit of electric current vi. SI unit of electric current |
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Answer» i. Centimetre Gram Second ii. Foot Pound Second iii. Metre (m) iv. Cd v. Kelvin (K) vi. Ampere (A) |
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| 18. |
How many molecules are present in one mole of substance? |
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Answer» 6.022x1023 molecules |
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| 19. |
Define molar volume of a gas. |
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Answer» It is defined as volume occupied by one mole of any substance. Molar volume of a gas= 22.4 L at STP (273 K, 1 atm) |
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| 20. |
When two or more atoms are combine in definite ratio is called:(A ) Mixture (B) Substance (C) compound (D) Formula . |
| Answer» (C) compound | |
| 21. |
Convert liter- atmosphere to joule (SI units of energy). |
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Answer» 1 L atm × 10-3 m3 /1L × 101.325Pa/atm =101.325 Pa m3 101.325 Pa m3 = 101.325(N/m2 ) m3 =101.325 N m= 101.325 J |
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| 22. |
How has chemistry contributed towards nation’s development? |
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Answer» Chemical principles are important in diverse areas such as weather patterns, functioning of brain, operation of a computer, chemical industries, manufacturing , fertilizers, alkalis, acids, salts, dyes, polymers, drugs, soaps, detergents, metals, alloys, contribute in a big way to national economy. |
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| 23. |
Give reason : The mass of a body is more fundamental property than its weight. |
Answer»
Hence, The mass of a body is more fundamental property than its weight. |
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| 24. |
The number of significant numbers in 0.0580 is;(A) 2 (B) 3 (C ) 4 (D ) 1 |
| Answer» (B) significant numbers is 3. | |
| 25. |
S. I Unit of mass is(A) Gram (B ) Milligram (C) Kilogram (D) Quintal |
| Answer» (C) Kilogram | |
| 26. |
28.7pm is equal to:(A ) 2.87x1012m (B) 2.87x10 -11m (C) 2.87x10-12 (D) 28.7x10-12. |
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Answer» (B) 2.87x10 -11m |
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| 27. |
How can we say that sugar is solid and water is liquid? |
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Answer» Sugar has close packing of constituent particles, have its own volume and shape therefore, it can be said to be solid whereas in water the constituent particles are not as closely packed as in solid. It has definite volume but not definite shape. Therefore it is a liquid. |
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| 28. |
Calculate the mass of one liter of mercury in grams and kilograms if the density of liquid mercury is 13.6 g cm– 3 . |
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Answer» \(\because\) Mass = Volume x Density Thus, mass of 1 liter of mercury (in gm) = 1 L\(\Big[\frac{1000\,cm^3}{1\,L}\Big]\) [13.6 g cm-3] = 1.36 × 104 g Mass in Kilogram = 1.36 × 104 g × \(\frac{1\,Kg}{100\,g}\) = 13.6 kg |
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| 29. |
How is gram related to the SI unit kilogram? |
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Answer» The SI unit kilogram (kg) is related to gram (g) as 1 kg = 1000 g = 103g. [Note : ‘Gram’ is used for weighing small quantities of chemicals in the laboratories. Other commonly used quantity is ‘milligram’. 1 kg = 1000 g = 106mg] |
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| 30. |
What is the effect of temperature on molarity of solution? |
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Answer» Molarity of a solution decreases on increasing temperature. |
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| 31. |
(i) What is limiting reactant? (ii) Oxygen is prepared by catalytic decomposition of potassium chlorate (KCIO3). Decomposition of potassium chlorate gives potassium chloride (KCI) and oxygen (O2). If 2.45 mol of oxygen is needed for an experiment, how many grams of potassium chlorate must be decomposed? |
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Answer» (i) It is the reactant which is entirely consumed when reaction goes to completion. (ii) 2KClO3(s) → 2KCl(s) + 3O2(g) Molecular weight of KClO3 = 39 + 35.5 + 3 × 16 = 122.5 For 3 moles of O2 we need = 2 × 122.5 g of KClO3 For 2.4 moles of O2 we need = \(\frac{2\times122.5}{3}\) x 2.4 = 196 g of KClO3. |
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| 32. |
Define Mole. |
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Answer» 1 mole represent the amount of substance which contain Avagadro's number of particles. |
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| 33. |
What is the mass of mercury in grams and in kilograms if the density of liquid mercury is 13.6 g cm-3? |
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Answer» Mass = Volume x density = 1000 cm3 × 13.6 g cm-3 =13,600 g = 13.6 kg |
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| 34. |
How can we say that sugar is solid and water is liquid? |
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Answer» Sugar is solid because it has definite volume and definite shape. Water is liquid because it has definite volume but not the definite shape. |
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| 35. |
What is limiting reactant in a reaction? |
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Answer» It is the reactant which gets consumed first or limits the amount of product formed. |
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| 36. |
The magnetic nature of elements depends on the presence of unpaired electrons. Identify the configuration of transition element, which shows highest magnetic moment. (i) 3d7 (ii) 3d5 (iii) 3d8 (iv) 3d2 |
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Answer» (ii) 3d5 shows highest magnetic moment. |
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| 37. |
Amino acid contains two functional groups, one is -COOH and other is (A) –OH (B) -NH3 (C) -NH2(D) NH4 |
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Answer» Correct Answer is: (C) -NH2 Amino acids get their name from two "functional groups" or groups of atoms common to all these compounds. One of these groups is a carboxylic acid, the (C=O) OH on each amino acid. |
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| 38. |
Define Avagadro’s Law. |
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Answer» Equal volume of all gases and vapours under the same condition of temperature and pressure contains equal number of molecules. |
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| 39. |
Calculate the number of moles of NaOH in 27 cm3 of 0.15 M NaOH solution. |
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Answer» M = n x \(\frac{1000}{Volume\,of\,solution\,in\,cm^3}\) 0.15 = n x \(\frac{1000}{27}\) n = \(\frac{0.15\times27}{1000}\) = 0.00405 moles |
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| 40. |
Calculate the mass percentage of hydrogen atom in 1 mole of water. |
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Answer» Molar mass of water = 18.02 g Mass % of hydrogen = \(\frac{Total\,mass\,of\,the\,elements\,in\,the\,compound}{Molar\,mass\,of\,the\,compound}\) x 100 = \(\frac{2\times1.008}{18.02}\) x 100 = 11.18% Mass % of oxygen = \(\frac{16.00}{18.02}\) x 100 = 88.79 |
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| 41. |
What is the relationship between molecular weight and V.D. of a gas? |
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Answer» Molecular Weight = 2 × V.D. |
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| 42. |
Give reasons for the following :(a) Compounds of transition elements are generally coloured. (b) MnO is basic while Mn2O7 is acidic. |
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Answer» (a) Due to d-d transition (b)Due to higher oxidation state of Mn2O7/Due to high polarizing power of Mn(VII). |
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| 43. |
Predict which of the following will be coloured in aqueous solution?Ti3+, V3+, Sc3+, Mn2+, Fe3+, Co2+ and Mnoa |
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Answer» Ions with incompletely filled d orbitals will be coloured. Due to d-d transition, Ti3+, V3+, Mn2+, Fe3+ and Co2+ are coloured. MnO4- is also coloured due to charge transfer. Only Sc3+ (3d0) is colourie. |
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| 44. |
Give reasons for the following :Calculate the magnetic moment of a divalent ion in aqueous medium if its atomic number is 26. |
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Answer» μ = √4(4+2) = 4.90 B.M . |
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| 45. |
Explain the following observations:Cr2+ is a stronger reducing agent than Fe2+ in aqueous solution. |
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Answer» Cr2+(d4) changes to Cr3+ (d3) while Fe2+ (d6) changes to Fe3+ (d5). In aqueous medium d3 is more stable than d5. |
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| 46. |
Give three main points of difference between a compound and a mixture. |
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| 47. |
Commercially available, concentrated hydrochloric acid contains 38% HCl by mass, (a) What is the molarity of this solution? The density is 1.19 g mL–1 (b) What volume of concentrated HCl is required to make 1.00 L of 0.10 M HCI? |
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Answer» (a) We know that, Molarity = \(\frac{Number\,of\,moles\,of\,solute(HCl)}{volume\,of\,solution}\) Number of moles of HCl = \(\frac{Mass\,of\,the\,solute(HCl)}{Molecular\,mass\,of\,HCl}\) = \(\frac{38}{36.5}\) Volume of solution: 38 g of HCl is present in 100 g of the solution Volume of 100 g of the acid = \(\frac{Mass}{Density}\) = \(\frac{100\,g}{1.19\,g\,cm^{-3}}\) [\(\because\) 1mL = 1cm3] = 84.033 ≈ 84.05 cm3 Molarity = \(\frac{\frac{38}{36.5}}{\frac{84.05}{1000}}\) = 12.38 M (b) For monobasic acid M1V1 = M2V2 12.38 M × V1 = 10 M × 1000 mL V1 = \(\frac{10\,M\times1000\,mL}{12.38\,M}\) = 807.754 ≈ 807.8 mL |
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| 48. |
If LPG cylinder contains a mixture of butane and isobutene, then the amount of oxygen that would be required for combustion of 1 kg of it will bea. 1.8 kgb. 2.7 kgc. 4.5 kgd. 3.58 kg |
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Answer» Correct option is d. 3.58 kg |
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| 49. |
The mass of 70% H2SO4 required for neutralization of 1 mol of NaOH isa. 49 gb. 98 gc. 70 gd. 34 g |
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Answer» Correct option is c. 70 g |
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| 50. |
30 mL of an acid is neutralized by 15 mL of 0.2 N alkali. The strength of the acid is a. 0.1 Nb. 0.2 Nc. 0.3 Nd. 0.4 N |
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Answer» Correct option is a. 0.1 N |
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