Some Basic Concepts of Chemistry Most Important Questions

Q1. A compound prepared by any method contains the same elements in the fixed ratio by mass. The given statement is known as:

• 1) Law of multiple Proportions
• 2) Law of conservation of mass
• 3) Law of reciprocal Proportions
• 4) Law of definite Proportions

Solution

According to the law of definite proportions, a given compound always contains exactly the same proportion of elements by mass.

Q2. The empirical formula of an organic gas containing carbon and hydrogen is CH₂. The mass of 1 litre of this organic gas is exactly equal to that of 1 litre of N₂. Therefore, the molecular formula of organic compound is:

• 1) C₆H₁₂
• 2) C₄H₈
• 3) C₂H₄
• 4) C₃H₆

Solution

Molecular mass of the gas = Molecular mass of N₂ = 28 g Molecular formula mass = n X (Empirical formula mass) Empirical formula mass of the gas= 12 + 2 = 14 g

 

Hence the molecular formula is (CH₂)₂ or C₂H₄.

Q3. What is the simplest formula of the compound which has the following percentage composition? Carbon 80% and Hydrogen 20%. If the molecular mass is 30, calculate its molecular formula.

Solution

Calculation of empirical formula: Element Percentage Atomic mass Relative number of moles Simple ratio Simplest whole number ratio C     H 80     20 12     1    1     3   Therefore Empirical formula is CH₃  

Calculation of molecular formula:  Empirical formula mass = 12 × 1 + 1 × 3 = 15

  

Molecular formula = Empirical formula × 2 = CH₃ × 2 = C₂H₆

Q4. A statement which is not a part of Dalton’s atomic theory is:

• 1) Matter is made of atoms.
• 2) Atoms are composed of sub-atomic particles called electron, proton and neutron.
• 3) All atoms of a given element are identical.
• 4) Atoms of different elements have different mass and different properties.

Solution

According to the Dalton’s atomic theory, atoms are indivisible i.e. they are not composed of any sub-atomic particles.

Q5. 18 g of glucose (C₆H₁₂O₆) is present in 1000 g of an aqueous solution of glucose. The molality of this solution is:

• 1) 0.5m
• 2) 1m
• 3) 1.1m
• 4) 0.1m

Solution

18 g of glucose (C₆H₁₂O₆) is present in 1000 g of an aqueous solution of glucose. The molality of this solution is 0.1 M. Solution: Given mass of Glucose = 18 g Mass of Solvent = 1000 g Molar mass of glucose = 180 g

 

Q6. Classify the following substances as elements, compounds and mixtures. In case of mixtures clearly indicate whether the mixture is homogenous or heterogeneous and also explain how would you separate them?

• a) Iron
• b) Salt and water
• c) Glass powder, iron filings and sugar
• d) Distilled water

Solution

a) Iron is an element

b) Salt and water are a homogenous mixture which can be separated by distillation, in this process salt remains as residue.

c) Glass powder, iron filing and sugar are a heterogeneous mixture. To separate the constituents dissolve the mixture in water, sugar dissolves. Now filter to separate undissolved glass powder and iron filings. Glass powder and iron filings can be separated with the help of a magnet. Sugar can be recovered from solution by evaporation.

d) Distilled water is a compound.

Q7. Gastric juice contains 3.0 g of HCl per litre. If a person produces about 2.5 litres of gastric juice per day, How many antacid tablets each containing 400 mg of Al(OH)3 are needed to neutralize all the HCl produced in one day.

Solution

Al(OH)₃  +  3HCl  →  AlCl₃  +  2H₂O   78 g                    109.5 g   

Q8. 10 ml of HCl solution produced 0.1435 g of AgCl when treated with excess of silver nitrate solution. What is the molarity of acid solution? (Atomic mass of Ag = 108).

Solution

The chemical equation of the reaction is:      HCl  +  AgNO₃   →    AgCl  +  HNO₃  36.5 g                     143.5g 

  

Q9. Calculate the number of moles of carbon dioxide which contain 8g of oxygen.

Solution

Q10. A non-metal X forms two oxides I and II. The mass percentage of oxygen in I (X₄O₆) is 43.7% which is same as that of X in the 2^nd oxide. What is the formula of 2^nd oxide?

Solution

Oxygen  X  I oxide 43.7  56.3 II oxide  56.3   43.7   From the data given above we can say that 43.7 parts of Oxygen corresponds to = 6 oxygen atoms

   

Also 56.3 parts of X in I correspond to = 4 X atom 

 

Hence, the formula of second oxide is X₂O₅

Q11. Normality is defined as:

• 1) Amount of solute present in 100g of solution
• 2) Number of moles of solute per litre of solution
• 3) Number of gram equivalents of solute per litre of solution
• 4) Number of moles of solute per 1000g of solvent

Solution

Normality of a solution is the number of gram equivalents of solute per litre of solution.

Q12. The number of significant figures in 0.04 are:

• 1) 3
• 2) 1
• 3) 2
• 4) 4

Solution

Zeros before the decimal point are not significant.  

Q13. Carbon forms two gaseous oxides. One of these oxides contains 42.8% carbon while the other contains 27.27% carbon. Explain which law of chemical combination this data illustrates. State this law.

Solution

The ratio of oxygen in the two compounds is 1:2 so this compound follows law of multiple proportions. According to this law when two elements combine with each other to form two or more than two compounds, the masses of one of the elements which combine with fixed mass of the other, bear a simple whole number ratio to the other.

Q14. If 28.036 is rounded off to 4 significant figures, then the result will be:

• 1) 28.03
• 2) 28.036
• 3) 28.0
• 4) 28.04

Solution

  28.04

Q15. Write a short note on formula unit mass.

Solution

Formula Mass Mass of a molecule of an ionic compound. It is  the sum of atomic masses of all the elements present in one formula unit of a compound. It is used for measuring the molecular mass of entities which do not exist in the solid form.

Substances such as sodium chloride do not contain discrete molecules as their constituent units. In such compounds, positive (sodium) and negative (chloride) entities are arranged in a three-dimensional structure.

In sodium chloride, one Na⁺ is surrounded by six Cl⁻ and vice versa. Hence, in such cases, the formula mass is used to calculate the mass instead of the molecular mass. Just like the molecular mass, it is calculated by adding the atomic masses of the atoms present in one formula unit.

Thus, formula mass of sodium chloride = atomic mass of sodium + atomic mass of chlorine       

 = 23.0 u + 35.5 u = 58.5 u

Q16. How many moles of electrons weight one kilogram?

(Mass of electron = 9.1 Χ 10^-31kg, avogadro number = 6.023 Χ 10 ²³)

• 1) 
• 2) 
• 3) 
• 4) 6.023 Χ 10²³

Solution

Given: Mass of one electron = 9.108 x 10-31 kg 

Q17. Number of moles of hydroxide ions in 0.3L of 0.005M solution of Ba (OH)₂ is:

• 1) 0.0075
• 2) 0.003
• 3) 0.0015
• 4) 0.0050

Solution

No of moles = Molarity x volume in litres = 0.005 × 0.3 = 0.0015 moles of Ba(OH)₂ 1mole of  Ba(OH)₂ yields 2 moles of hydroxide ions

Q18. What are significant figures? Give the number of significant figures in each of the following?

• a) 2.56 Χ 10³ 
• b) 165
• c) 2.05
• d) 5000
• e) 0.00256

Solution

The number of significant figures in a measurement is the number of figures that are known with certainty plus one that is uncertain, beginning with one non-zero digit.

a) Number of significant figures in 2.56 Χ 10³= 3

b) Number of significant figures in 165 = 3 (all non-zero digits are significant).

c) Number of significant figures in 2.05 = 3 ( zero’s between non zero digits are significant)

d) Number of significant figures in 5000 = 1 (if a number ends in zero’s that are not to the right of a decimal, the zeros may or may not be significant.)

e) Number of significant figures in 0.00256 = 3 (zero’s to the left of the first non-zero digit in the number are non significant.)

Q19. A chloride of phosphorus contains 22.57% P. Phosphine contains 8.82% hydrogen and hydrogen chloride gas contain 97.26% chlorine. Show that the data illustrates law of reciprocal proportions:

Solution

i) In phosphorus chloride, P = 22.57 %,        Cl = 77.43%  

ii) In phosphine,   P = 91.18%, H = 8.82% 

 

Thus, ratio of the masses of Cl and H which combine with fixed mass (22.57 parts) of phosphorus separately is, 77.43: 2.18   or   35.5: 1  

iii) In hydrogen chloride   Cl = 97.26%         H = 2.77%   Thus, ratio of masses of Cl and H when they combine with each other is   97.26: 2.77        or    35.5: 1  

iv) Ratio 1: Ratio 2 =  This shows a simple whole number ratio.

Q20. 1.375 g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper that remained was 1.098g. In another experiment 1.179 g of copper was dissolved in nitric acid and the resulting copper nitrate was converted into cupric oxide by ignition .the weight of cupric oxide formed was 1.476 g. which law of chemical combinations does this data state?

Solution

In first experiment: Copper oxide = 1.375 g Copper left = 1.098 g Therefore oxygen present = 1.375 – 1.098g = 0.277 g

  

Second experiment: Copper taken = 1.179 g Copper oxide formed = 1.476 g Therefore oxygen present = 1.476-1.179g = 0.297 g 

 

Since percentage of oxygen is the same in both the above cases, so the law of constant composition is illustrated.

Q21. What is the mass (mg) of a zinc block whose dimensions are 2.0″ Χ  3.0″ Χ 5.0″ and whose density is 2.5g /cm³? (Given 1.0″= 2.54 cm.)

Solution

Total volume of a zinc block = 2.0 Χ 3.0 Χ 5.0 = 76.20 Hence mass in grams = density Χ volume = 2.5 Χ 76.20 = 190.500 g = 1.905 Χ 10² g

Q22. If 0.5 mole of BaCl₂ is mixed with 0.2 mole of Na₃PO_4, then the maximum number of moles of Ba₃(PO₄)₂ that can be formed will be:

• 1) 0.5
• 2) 0.3
• 3) 0.1
• 4) 0.7

Solution

3BaCl₂  +  2Na₃PO₄  → Ba₃(PO₄)₂ + 6NaCl From the molar ratio, we get, BaCl₂  :  2Na₃PO₄  :  Ba₃(PO₄)₂  : 6NaCl   3: 2:1:6 Also, limiting reactant is Na₃PO₄ Therefore, 0.2 mol Na₃PO₄ will give Ba₃(PO₄)2 = 0.5 Χ 0.2 = 0.1 mol.

Q23. According to Dalton’s atomic theory chemical reactions involve:

• 1) reorganisation of nuclei
• 2) destruction of atoms
• 3) reorganisation of atoms
• 4) construction of atoms

Solution

According to Dalton’s atomic theory chemical reactions involve reorganisation of atoms.

Q24. Which of the following is an element?

• 1) Sugar
• 2) Air
• 3) Oxygen
• 4) Urea

Solution

Sugar and urea are compounds while air is a mixture. Oxygen is an element having symbol O and exists as diatomic molecule O₂.

Q25. What is the molecular mass of a compound if 20% nitrogen is present in it?

Solution

  Q26. The molecular mass of methane and oxygen are 16 and 32 respectively. If one litre of methane at S.T.P contains N molecules, what will be the number of molecules in 5 L of oxygen at S.T.P?

Solution

According to Avogadro’s law equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules. So 5 L of oxygen at S.T.P contains 5N molecules. 

Q27. Express the following in SI units:

• A) 93 million miles
• B) 5 feet 2 inches  

Solution

A) 1 mile = 1.6 Χ 10³ m Therefore, 93 million miles = 93 Χ 10⁶ Χ 1.60 Χ 10³ m = 1.49 Χ 10¹¹ m

B) 5 feet 2 inches = 5 Χ 12 + 2 = 62 inches = 62 Χ 2.54 Χ 10^-2 m = 157.48 Χ 10^-2 m = 1.5748 m

Q28. What volume of 10M HCl should be diluted with water to prepare 2.00L of 5M HCl?

Solution

In case of dilution,     M₁V₁ =  M₂V₂ 10 M HCl 5 M HCl   10 × V₁ = 5 × 2.00   

Q29. Two acids H₂SO₄ and H₃PO₄ are neutralized separately by the same amount of an alkali when sulphate and dihydrogen orthophosphate are formed respectively. Find the ratio of masses of H₂SO₄ and H₃PO₄ [at mass P = 31] 

Solution

H₂SO₄ + 2NaOH → Na ₂SO₄ + 2H₂O   H₃PO₄ + NaOH  → NaH₂PO₄  +  H₂O   Equivalent of alkali = 1 g equivalent of H₂SO₄ = 1 g equivalent of H₃PO₄ The two acids must react in the ratio of their equivalent masses.

  

Hence ratio of masses of H₂SO₄ : H₃PO₄ = 49 : 98 =1 : 2.

Q30. What is the mass of 2 litre sample of water containing 25% heavy water (D₂O) in it by volume? Density of water is 1.0 g cm^-3 where as that of D₂O is 1.06 g cm^-3.

Solution

Q31. How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?

Solution

0.50 mol Na₂CO₃ means, 0.50 Χ 10⁶ = 53 g of Na₂CO₃ where as 0.50 M Na₂CO₃ means 1 litre solution of sodium carbonate contains 0.5 moles of sodium carbonate.

Q32. What were the limitations of Dalton’s atomic theory? Explain in detail.

Solution

The main failures of Dalton atomic theory are:

(1) It failed to explain how atoms of different elements differ from each other i.e. It did not tell anything about structure of the atom.

(2) It could not explain how and why atoms different elements combine with each other to form compound atoms or molecules.

(3) It failed to explain the nature of forces that bind together different atoms in a molecule.

(4) It failed to explain Gay Lussac’s law of combining volumes.

(5) It did not make any distinction between ultimate particle of an element that takes part in reactions (atoms) and ultimate particle that has independent existence (molecule).

Q33. 15g of a substance A combines with 20g of a substance B to give 35 g of product C. The law followed in this reaction is:

• 1) Law of conservation of mass
• 2) Law of reciprocal proportions
• 3) Law of definite proportions
• 4) Law of multiple proportions

Solution

According to the law of conservation of mass, matter can neither be created nor destroyed in a chemical reaction. It always remain conserved.

Q34. Balance the following equations:

• i) Fe + H₂O → Fe₃O₄ + H₂ 
• ii) KMnO₄ + H₂SO₄ → K₂SO₄ + MnSO₄ + H₂O + O₂ 
• iii) I₂ + HNO₃ → HIO₃ +NO₂ + H₂O
• iv) Zn + HNO₃ → Zn(NO₃)₂ + N₂O + H₂O  

Solution

i) 3Fe +4 H₂O → Fe₃O₄ + 4H₂ 

ii) 4KMnO₄ + 6 H₂SO₄ → 2K₂SO₄ + 4MnSO₄ + 6H₂O + 5O₂ 

iii) I₂ + 10HNO₃ → 2HIO₃ + 10 NO₂ + 4H₂O

iv) 4Zn + 10 HNO₃ → 4Zn (NO₃)₂ + N₂O + 5H₂O

Q35. Explain Law of Combining Volumes in detail.

Solution

Gay-Lussac’s Law of Gaseous Volumes (Law of Combining Volumes) This law was given by Gay-Lussac in 1808. He investigated a large number of chemical reactions occurring in gases. As a result of his experiments, Gay-Lussac found that there exists a definite relationship among the volumes of gaseous reactants and products.

Consider the following illustrations. When gases react together to produce gaseous products, the volumes of reactants and products bear a simple whole number ratio with each other, provided volumes are measured at same temperature and pressure.

Example 1: Under identical conditions of temperature and pressure equal volumes, say 1 litre i.e., 1L of each hydrogen and chlorine gases react together to produce double the volume, 2L of hydrogen chloride gas.

Thus, the ratio of volumes is 1 : 1 : 2.

Example 2: Under identical conditions of temperature and pressure 2l of hydrogen gas reacts with 1L of oxygen gas to produce 2L of steam (water vapour)

Thus, the ratio of volumes is 2 : 1 : 2.

Example 3: Under identical conditions of temperature and pressure 1L of nitrogen gas reacts with 3L of hydrogen to produce 2L of ammonia gas.

Here the ratio of volumes is 1 : 3 : 2. Conclusion: All

Examples 4 cited above indicate that during the gaseous reaction, gaseous reactants and products bear a simple ratio of a whole number of volumes with each other.

Q36. How many grams of oxygen (O₂) are required to completely react with 0.200 g of hydrogen (H₂) to yield water (H₂O)?  

Solution

Q37. How many litres of oxygen are required for the complete combustion of 125 litres of ethylene (C₂H₂)?

Solution

2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O 5 Moles of O₂ is required for the combustion of 2 moles of C₂H_2.

Q38. Sapphire weighs 563 carats. If one carat is equal to 200mg, what is the weight of the gemstone in grams?

Solution

Weight of one carat = 200 mg 

Q39. 2.5 moles of sulphuryl chloride were dissolved in water to produce sulphuric acid and hydrochloric acid. How many moles of KOH will be required to completely neutralize the solution?

Solution

SO₂Cl₂ + 2H₂O → H₂SO₄ + 2HCl Mol of HCl produced from 2.5 moles of SO₂Cl₂ = 5 mol   Mol of H₂SO₄ produced from 2.5 moles of SO₂Cl₂ = 2.5 mol   H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O;   HCl + KOH → KCl + H₂O   Now, 5 mol of H₂SO₄ require KOH = 2.5 mol   Total KOH required = 10 + 2.5 = 12.5 mol. 

Q40. Calculate the molecular formula of a gaseous compound whose 1 volume requires 2 volumes of O₂ for combustion and gives 2 volumes of CO₂ and 1 volume of N₂ after it.

Solution

According to the question,X + 2O₂ → 2CO₂ + N₂ Since the product has 2 atoms each of C and N more than that of the reactant, these must be present in the X (reactant). Hence X is C₂N₂.