1

Q1. Nitrogen monoxide reacts with oxygen according to the reaction: 2NO (g) + O₂ (g) → 2NO₂(g) If 12 moles of nitrogen monoxide reacts with 10 moles of oxygen, how many moles of NO₂ will be formed?  

• 2 moles
• 8 moles
• 6 moles
• 12 moles

Solution

NO is the limiting reagent hence, 12 moles of NO₂ will be formed by 12 moles of NO.

Q2. Which one of the following pair of substances illustrate the law of multiple proportions?

• KCl and KBr
• BaO and Ba(OH)₂
• SO₂ and SO₃
• H₂ and D₂

Solution

The law of multiple proportion states that, ''If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers.''

Q3. What is the empirical formula of C₂H₂ and C₆H₆?

• CH
• C₂H₂
• C₃H₃
• C₆H₆

Solution

Empirical formula represents the simplest whole number ratio of various atoms in a compound hence for both C₂H₂ and C₂H₆ the empirical formula is CH.

Q4. If empirical formula of a compound is C₅H₄ and the value of n is 2, then molecular formula of the compound is:

• C₅H₄
• C₂H₄
• C₁₀H₈
• C₁₅H₁₂

Solution

Molecular formula of the compound is (C₅H₄)_n since here n = 2 therefore molecular formula will be C₁₀H₈.    

Q5. What is the empirical formula of a compound which contains 2.63 moles of N and 3.947 moles of O?  

• NO
• NO₂
• N₂O₅
• N₂O₃

Solution

Mole ratio N : O is 2.63 : 3.947 = 1 : 1.5 Therefore the empirical formula is N₂O₃.  

Q6. Which of the following compounds has the same empirical formula as that of glucose (C₆H₁₂O₆)?

• CH₃CHO
• CH₃COOH
• CH₃OH
• C₂H₆

Solution

The empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound. Molecular formula of glucose is C₆H₁₂O₆. Its empirical formula is CH₂O. For CH₃COOH, the empirical formula is also CH₂O.

Q7. If you have 2 moles of A and 3 moles of B, Identify the limiting reagent in the following reaction. A + B →  AB

• A
• B
• A as well as B
• None

Solution

2 moles of A will react with 2 moles of B and 1 mole of B will be left unreacted.

Q8. The coefficient in 6.65 x 10⁴ is:

• 10⁴
• 6.65
• 6.65 x 10⁴
• 10^-4

Solution

Here 6.65 is the coefficient and 10⁴ is the exponent.

Q9. Number of nitrogen atoms present in 1.4 g of N₂:

• 6.023 X 10²³
• 3.012 X 10²³
• 6.023 X 10²²
• 3.012 X 10²²

Solution

  One mole of N₂ molecule contains 6.023 X 10²³ moecules of N₂ and each N₂ molecule contains 2 atoms of N.  

Q10. An oxide of metal M has 40% by mass of oxygen. Metal M has relative atomic mass of 24g/mol. The empirical formula of the oxide is:

• M₂O
• M₂O₃
• MO
• M₂O₅

Solution

Hence, ratio of moles of M and O = 1:1 Therefore, the formula of the compound is MO.

Q11. If 2 moles of H₂ and O₂ are available for the reaction, identify the limiting reagent for the reaction, 2H₂ + O₂ → 2H₂O  

• H₂
• O₂
• H₂O
• Both H₂ and O₂

Solution

Only 1 mole of O₂ is required for the complete combustion of 2 moles of H_2, hydrogen is the limiting reagent.

Q12. Law of conservation of mass was given by:

• Antoine Lavoisier
• Avogadro
• Joseph Proust
• John Dalton

Solution

Antoine Lavoiser gave the law of conservation of mass.  

Q13. If the true value for a result is 3.00 m and a student records two readings as 3.01 m and 2.99 m, then we can conclude that:

• the values are accurate
• the values are precise
• the values are both accurate and precise
• the values are neither accurate nor precise

Solution

Precision indicates how closely repeated measurements match each other. Accuracy indicates how closely a measurement matches the correct or expected value.

Q14. Molar mass of F₂ is 38 g. How many atoms are present in 0.147 mole of F₂?

• 6.02 Χ 10²³ atoms
• 0.292 atoms
• 8.78 Χ 10²² atoms
• 1.76 Χ 10²³ atoms

Solution

One molecule of F₂  contains 2 atoms of fluorine. 1 mol of fluorine contains 2 X 6.023 X 10²³ atoms So, 0.147 moles will contain 0.147 X 2 X 6.023 X 10²³ = 1.76 X 10²³ atoms of flourine.  

Q15. Amount of water produced by the combustion of 16g of methane is:

• 36 g
• 32 g
• 34 g
• 30 g

Solution

                                  According to the above equation, one mole of CH₄ (16 g) on combustion gives one mole of carbon dioxide and 2 moles of water and 2 moles of water = 2 x [2 + 16] = 36 g of water.

Q16. Two different atoms A and B combine together to form AB. What will be AB?

• Ion
• Element
• Atom
• Molecule

Solution

Two different atoms combine together chemically to form a molecule.

Q17. At what conditions Gay Lussac's law of gaseous volumes is considered?

• Same amount of all the reactants and products only.
• Same temperature and pressure.
• Same pressure only
• Same temperature only

Solution

Gay Lussac's law of gaseous volumes states that when gases combine or are produced in a chemical reaction, they do so in a simple ratio by volume provided all gases are at same temperature and pressure.  

Q18. The empirical formula of a compound is CH₂Cl. If the molecular mass of the compound is 99 g, what is the molecular formula of the compound?  

• C₂H₄Cl
• C₂HCl₂
• CH₄Cl₂
• C₂H₄Cl₂

Solution

Empirical formula mass = 49.5 g. Molecular mass = 99 g Therefore n = 2 Therefore molecular formula of compound = (CH₂Cl)₂

Q19. Amount of carbon dioxide produced when one mole of carbon is burnt in air is:

• 40g
• 44g
• 46g
• 20g

Solution

C(s) +O₂ (g)  CO₂(g) In this reaction 12g of carbon i.e 1 mole of C burns to produce 44g of carbon dioxide.

Q20. If the true value for a result is 2.00 g and a student takes 2 measurements and reports the results as 1.95 g and 1.93 g, then we can conclude that:

• The values are precise
• The values are accurate
• The values are both precise and accurate
• The value are neither precise nor accurate

Solution

Precision indicates how closely repeated measurements match each other. Since both the values are close to each other, they are precise and since their is a difference between the mean value and true value, they are not accurate.

Q21. Femto stands for:

• 10^-6
• 10^-15
• 10^-14
• 10^-12

Solution

Femto is a prefix used in SI system for the multiple 10^-15 and its symbol is f.

Q22. A compound has an empirical formula of C₂HBr. If the molar mass of the compound is 314.7 g/mol, what is the molecular formula of the compound? (Atomic mass of Br = 80g/mol)

• C₂HBr
• C₄H₂Br₂
• C₆H₃Br₃
• C₈H₄Br₄

Solution

Empirical formula mass is (2 X 12) + 1 + 80 = 105g and molar mass of the compound is 314 g/mol. Hence, n = 3 Therefore, the molecular formula is C₆H₃Br₃.  

Q23. Normality is defined as:

• Number of moles of solute per litre of solution
• Number of gram equivalents of solute per litre of solution
• Number of moles of solute per 1000g of solvent
• Amount of solute present in 100g of solution

Solution

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

Q24. According to Avogadro's law at the same temperature and pressure:

• Equal volumes of gases contain equal number of molecules.
• Matter can neither be created nor destroyed.
• Matter consists of indivisible atoms.
• Compounds are formed when atoms of different elements combine in a fixed ratio.

Solution

According to Avogadro's law, equal volumes of gases at same temperature and pressure should contain equal number of molecules.

Q25. 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:

• 0.7
• 0.5
• 0.3
• 0.1

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.

Q26. Atomic mass of Cl is 35.5 g. Calculate the mass of 4.50 moles of chlorine gas, Cl₂.

• 6.34 g
• 319.5 g
• 4.5 g
• 160 g

Solution

Mass of 1 mole of Cl₂ is 71 g. Hence mass of 4.50 moles is 71 X 4.50 = 319.5 g  

Q27. The number of moles of Cr and O in a compound are 4.8 x 10¹⁰ and 9.6 x 10¹⁰ respectively. Its empirical formula is:

• Cr₂O₃
• CrO₂
• Cr₂O₄
• Cr₃O₆

Solution

Molar ratio of Cr and O = 4.8 x 10¹⁰ : 9.6 x 10¹⁰ = 1 : 2 Hence, empirical formula = CrO₂

Q28. 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:

• Law of multiple proportions
• Law of definite proportions
• Law of conservation of mass
• Law of reciprocal 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.

Q29. Empirical formula of an organic compound is CH and its molecular mass is 78g/mol. What is its molecular formula?

• C₂H₂
• C₂H₄
• C₆H₆
• C₂H₆

Solution

Empirical formula of mass CH = (12 + 1) = 13 Molecular formula = Empirical formula x n

Q30. What is the empirical formula of C₄H₈O₂?

• C₂H₄O
• CHO
• C₄H₈O₂
• C₂H₂O₂

Solution

Empirical formula represents the simplest whole number ratio of various atoms in a compound hence here n=2 and the empirical formula is C₂H₄O.

Q31. The atomic mass of an element is usually fractional because:

• Elements contain impurities
• Elements are mixture of allotropes
• Elements are mixture of isotopes
• Elements are mixture of isobars

Solution

The atomic mass is the average of atomic masses of all the isotopes of an element.

Q32. The number of significant figures in 3.1205 are:

• 5
• 6
• 3
• 7

Solution

All non-zero digits are significant and the zeros between the non-zero digits are significant.

Q33. 18 g of glucose (C₆H₁₂O₆) is present in 1000 g of an aqueous solution of glucose. The molality of this solution is:

• 1m
• 1.1m
• 0.5m
• 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

Q34. The simplest formula of a compound containing 50% of element X (atomic mass = 10g/mol) and 50% of element Y (atomic mass 20g/mol) is:

• XY
• X₂Y
• XY₃
• X₂Y₃

Solution

Q35. The scientific notation for 700 is:

• 7 x 10²
• 7 x 10^-2
• 7 x 10
• 7 x 10³

Solution

For expressing a quantity in scientific notation, we multiply it by 10 raised to the power of a positive number if it is a number larger than. 7 x 10² = 700

Q36. Volume of 17g of NH₃ at N.T.P. will be:

• 22.4 L
• 2.24 L
• 44.8 L
• 4.48 L

Solution

Molar mass of NH₃ is 17g. So the volume of 17g of NH₃ i.e 1 mole NH₃ at N.T.P. will be 22.4L.

Q37. The empirical formula of an acid CH₂O₂ is:

• CH₂O
• CH₂O₂
• C₂H₄O₂
• C₃H₆O₄

Solution

Since the formula of CH₂O₂ cannot be reduced to a simpler formula which represents the whole number ratio of various atoms in a compound, the empirical formula is same as molecular formula.

Q38. Substances whose two or more components completely mix with each other to make a uniform composition are called:

• Heterogeneous mixtures
• Homogenous mixtures
• Ionic compounds
• Elements

Solution

Homogenous mixtures are those whose components completely mix with each other to make a uniform composition.  

Q39. Which of the following has more number of atoms?

• 1 mole of Oxygen gas
• 1 mole of Ozone gas
• 1 mole of Neon gas
• 1 mole of Carbon monoxide gas

Solution

Since atomicity of Ozone (3) is maximum so, it will contain maximum number of atoms.

Q40. What is the mass percentage of carbon in urea [CO(NH₂)₂]? (Molar mass of urea is 60g)

• 2%
• 20%
• 32%
• 21%