Some Basic Concepts of Chemistry Important Questions Answers

Class 11 Some Basic Concepts of Chemistry Important Questions with Answers

Short Answer Type Questions

Q1. What will be the mass of one atom of C-12 in grams?

Answer.

The mass of 1 mole of C-12 atoms = 12 g

1 mole of C-12 atoms = 6.022 × 10²³ atoms

The mass of 1 atom of C-12 = 12 / (6.022 × 10²³)

= 1.99 × 10^–23 g

Q2. How many significant figures should be present in answer to the following calculation?

\(\begin{array}{l}\frac{2.5\times 1.25\times 3.5}{2.01}\end{array} \)

Answer.

The number of significant figures that should be present in the calculation

\(\begin{array}{l}\frac{2.5\times 1.25\times 3.5}{2.01}\end{array} \)

is 2.

Q3. What is the symbol for the SI unit of the mole? How is the mole defined?

Answer.

The symbol for the SI unit of the mole is mol.

One mole is defined as the amount of a substance containing the same number of particles or entities as there are atoms in exactly 12 g (0.012 kg) of the C – 12 isotope.

Q4. What is the difference between molality and molarity?

Answer.

Q5. Calculate the mass percent of calcium, phosphorus and oxygen in calcium phosphate Ca₃(PO₄)₂

Answer.

Molecular formula of calcium phosphate is Ca₃(PO₄)₂

Its molar mass will be – 3(40) + 2(31) + 8(16) = 310 g/mol.

\(\begin{array}{l}Percentage\ of\  Calcium = \frac{120}{310}\times 100 \end{array} \)

Percentage of Calcium = 38.71%

• Mass percent of phosphorus

\(\begin{array}{l}Percentage\ of\ Phosphorus = \frac{62}{310}\times 100\end{array} \)

Percentage of Phosphorus =20%

\(\begin{array}{l}Percentage\ of\ Oxygen= \frac{128}{310}\times 100\end{array} \)

Percentage of Oxygen = 41.29%

Q6. 45.4 L of dinitrogen reacted with 22.7 L of dioxygen and 45.4 L of nitrous oxide was formed. The reaction is given below:

2N₂(g) + O₂(g) → 2N₂O(g)

Which law is being obeyed in this experiment? Write the statement of the law.

Answer.

The volumes of dinitrogen and dioxygen that combine (i.e., 45.4 L and 22.7 L) have a simple ratio of 2: 1. As a result, it follows Gay Lussac’s law of gaseous volumes.

According to this law, “when gases combine or are produced in a chemical reaction, they do so in a simple volume ratio provided all gases are at the same temperature and pressure.”

Q7. 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 a whole-number ratio.

1. Is this statement true?
2. If yes, state according to which law?
3. Give one example related to this law.

Answer.

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 a whole-number ratio.

1. Yes, the statement is true.
2. According to the law of multiple proportions.
3. Consider the example,
   H₂ + O₂ → H₂O
   H₂ + O₂ → H₂O₂

Here masses of oxygen, (i.e., 16g in H₂O and 32g in H₂O₂) which combine with a fixed mass of hydrogen (2g) are in the simple ratio of 16 : 32 or 1 : 2.

Q8. Calculate the average atomic mass of hydrogen using the following data:

Answer.

The average atomic mass is given by the following formula-

\(\begin{array}{l}\frac{Natural\ Abundance\ Of _{}^{1}\textrm{H}\times Molar\ Mass+Natural\ Abundance\ Of\ _{}^{2}\textrm{H}\times Molar\ Mass}{100}\end{array} \)

\(\begin{array}{l}\frac{(99.985\times 1)+\left ( 0.015\times 2 \right )}{100}=1.00015\upsilon \end{array} \)

Hence, the average atomic mass will be 1.00015 μ.

Q9. Hydrogen gas is prepared in the laboratory by reacting dilute HCI with granulated zinc.

The following reaction takes place.

Zn + 2HCl → ZnCl₂ + H₂

Calculate the volume of hydrogen gas liberated at STP when 32.65 g of zinc reacts with HCI. 1 mol of a gas occupies 22.7 L volume at STP; atomic mass of Zn = 65.3 u.

Answer.

65.3 g of Zn reacts with HCl to form 22.7 of H₂ gas.

∴ 32.65 g of Zn at STP reacts with HCl to form =

\(\begin{array}{l}\frac{22.7\times 32.65}{65.3}=11.35 L\end{array} \)

Q10. The density of the 3 molal solution of NaOH is 1.110 g mL^–1. Calculate the molarity of the solution.

Answer.

Molality is 3 m

\(\begin{array}{l}Molality=\frac{number\ of\ moles\ of\ solute}{mass\ of\ solvent}\end{array} \)

\(\begin{array}{l}3=\frac{n}{mass\ of\ solvent}\end{array} \)

\(\begin{array}{l}mass\ of\ solvent=\frac{1000}{3}\end{array} \)

Also, number of moles=mass/molar mass

For NaOH,

\(\begin{array}{l}n=\frac{m}{40}\end{array} \)

Density of solution ⍴ = 1.110 g/mol

Mass of 1 mL solution = 1.11g = m_solute + m_solvent

n = 2.97 × 10^–3 mol

\(\begin{array}{l}M=\frac{2.97\times 10^{-3}}{0.001}=2.97 M\approx 3M\end{array} \)

Q11. Volume of a solution changes with change in temperature, then, will the molality solution be affected by temperature? Give reason for your answer.

Answer.

The temperature has no effect on the molality of the solution because molality is expressed in mass, and mass remains constant as temperature changes.

Q12. If 4 g of NaOH dissolves in 36 g of H₂O, calculate the mole fraction of each component in the solution. Also, determine the molarity of solution (specific gravity of solution is 1g ml^-1)

Answer.

Mass of NaOH = 4 g

Number of moles of NaOH = 4g/40g = 0.1 mol

Mass of H₂O = 36 g

Number of moles of H₂O = 36g/18g = 2 mol

\(\begin{array}{l}Mole\ fraction\ of\ water = \frac{Number\ of\ moles\ of\ H_{2}O}{Number\ of\ moles\ of\ water+Number\ of\ moles\ of\ NaOH}\end{array} \)

\(\begin{array}{l}Mole\ fraction\ of\ water =\frac{2}{2+0.1}=\frac{2}{2.1}=0.95\end{array} \)

\(\begin{array}{l}Mole\ fraction\ of\ NaOH= \frac{Number\ of\ moles\ of\ NaOH}{Number\ of\ moles\ of\ water+Number\ of\ moles\ of\ NaOH}\end{array} \)

\(\begin{array}{l}Mole\ fraction\ of\ NaOH=\frac{0.1}{2+0.1}=\frac{0.1}{2.1}=0.047\end{array} \)

Mass of solution = mass of water + mass of NaOH = 36g+4 g = 40 g

Volume of solution =40×1=40 mL.( Since specific gravity of solution is =1 g mL^–1)

\(\begin{array}{l}Molarity=\frac{number\ of\ moles\ of\ solute}{Volume\ of\ solution(V)}\end{array} \)

\(\begin{array}{l}Molarity=\frac{0.1\ mol\ NaOH}{0.04L}=2.5M\end{array} \)

Q13. The reactant which is entirely consumed in the reaction is known as a limiting reagent.

In the reaction 2A + 4B → 3C + 4D, when 5 moles of A react with 6 moles of B, then

(i) Which is the limiting reagent?

(ii) Calculate the amount of C formed?

Answer.

2A + 4B ⟶ 3C + 4D

According to the above equation, 2 moles of ‘A’ require 4 moles of ‘B’ for the reaction. As a result, the moles of ‘B’ required for 5 moles of ‘A’ are 10 moles.

(i) The limiting agent is B, as 5 moles of A requires 10 moles of B but only 6 moles are present.

(ii) The amount of ‘C’ formed will be determined by the amount of ‘B’ formed. Since 4 moles of ‘B’ produce 3 moles of ‘C’. As a result, 6 moles of ‘B’ will produce

\(\begin{array}{l}\frac{6\times 3}{4}=4.5\ moles\end{array} \)

Matching Type Questions

Q1. Match the following:

Answer.

Q2. Match the following physical quantities with units

Answer.

Assertion and Reason Type Questions

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Q1. Assertion (A): The empirical mass of ethene is half of its molecular mass.

Reason (R): The empirical formula represents the simplest whole-number ratio of various atoms present in a compound.

(i) Both A and R are true and R is the correct explanation of A.

(ii) A is true but R is false.

(iii) A is false but R is true.

(iv) Both A and R are false.

Answer.

Correct Option is (i) Both A and R are true and R is the correct explanation of A.

Molecular Formula = n × Empirical formula

\(\begin{array}{l}n=\frac{Molecular\ Mass}{Empirical\ Formula\ Mass}\end{array} \)

Empirical formula of Ethene = C₂H₄

Empirical Formula Mass = 14 amu= ½ Molecular Mass of Ethene

The ratio of Carbon and Hydrogen in the empirical formula is 1: 2.

Q2. Assertion (A): One atomic mass unit is defined as one-twelfth of the mass of one carbon-12 atom.

Reason (R): Carbon-12 isotope is the most abundant isotope of carbon and has been chosen as the standard.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

Answer.

Correct Option is (i) Both A and R are true and R is the correct explanation of A.

Since C-12 is used as the standard atom, one atomic mass unit is defined as one-twelfth of the mass of one carbon – 12 atom. This is due to the fact that it has an equal number of protons and neutrons (6) and makes up the majority of matter.

Carbon-12 is the most abundant isotope of carbon.

Q3. Assertion (A): Significant figures for 0.200 are 3 whereas for 200 it is 1.

Reason (R): Zero at the end or right of a number is significant provided they are not on the right side of the decimal point.

(1) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not a correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

Answer.

Correct Option is (iii) A is true but R is false.

Zero at the end or to the right of a number is significant if it is on the right side of the decimal point. For example, 0.200 has 3 significant figures.

Q4. Assertion (A): Combustion of 16 g of methane gives 18 g of water.

Reason (R): In the combustion of methane, water is one of the products.

(i) Both A and R are true but R is not the correct explanation of A.

(ii) A is true but R is false.

(iii) A is false but R is true.

(iv) Both A and R are false.

Answer.

Correct Option is (iii) A is false but R is true.

CH₄ + 2O₂ → CO₂ + 2H₂O

Water is produced during the combustion of methane, but 16 g of methane on complete combustion gives 36 g of water.

Long Answer Type Questions

Q1. A vessel contains 1.6 g of dioxygen at STP (273.15K, 1 atm pressure). The gas is now transferred to another vessel at a constant temperature, where pressure becomes half of the original pressure. Calculate

(i) volume of the new vessel.

(ii) number of molecules of dioxygen.

Answer.

(i) P₁ = 1atm, P₂ = ½ = 0.5 atm, T₁ = 273.15, V₂ = ?, V₁ = ?

32 g of dioxygen occupies = 22.4 L volume at STP

∴ 1.6 g of dioxygen will occupy =

\(\begin{array}{l}\frac{22.4\times 1.6g}{32g}=1.12 L\end{array} \)

V₁ = 1.12 L

From Boyle’s law (as T is constant)

_P1V₁ = P₂V₂

\(\begin{array}{l}V_{2}=\frac{1 atm\times 1.12 L}{0.6 atm}= 2.24 L\end{array} \)

V₂ = 22.4 L

(ii) Number of moles of dioxygen = Mass of dioxygen/Molar mass of dioxygen

\(\begin{array}{l}Number\ of\ moles\ of\ dioxygen = \frac{1.6}{3.2}=0.05\ mol\end{array} \)

1 mol of dioxygen = 6.022 × 10²³ molecule of dioxygen

∴ 0.05 mol of dioxygen = 6.022 × 10²³ × 0.5 molecules of dioxygen

= 0.3011 × 10²³ molecules

= 3.011 × 10²² molecules.

Q2. Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction given below:

CaCO₃ (s) + 2HCl (aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)

What mass of CaCl₂ will be formed when 250 mL of 0.76 M HCI reacts with 1000 g of CaCO₃? Name the limiting reagent. Calculate the number of moles of CaCl2 formed in the reaction.

Answer.

Number of moles of HCl =

\(\begin{array}{l}250\ mL \times \frac{0.76 M}{1000}= 0.19\ mol\end{array} \)

Mass of CaCO₃ = 1000 g

Number of moles of CaCO₃ = 1000 g/100g = 10 mol

According to the equation, 1 mol of CaCO₃ requires 2 mol of HCl.

Hence, for the reaction of 10 mol of CaCO₃ number of moles of HCl required would be:

\(\begin{array}{l}10\ mol\ CaCO_{3}\times \frac{2mol\ HCl}{1mol\ CaCO_{3}}=20\ mol\ HCl\end{array} \)

There is only 0.19 mol of HCl available, hence HCl is a limiting reagent.

Therefore, the amount of CaCl₂ formed will depend on the amount of HCl would give-

\(\begin{array}{l}0.19\ mol\ of\ HCl\times \frac{1mol\ CaCl_{2}}{2mol\ HCl}=0.095\ mol\ CaCl_{2}\end{array} \)

Or 0.095 × molar mass of CaCl₂ = 0.095 × 111 = 10.54 g.

Q3. Define the law of multiple proportions. Explain it with two examples. How does this law point to the existence of atoms?

Answer.

Dalton first studied the law of multiple proportions in 1803, and it can be stated as follows.

When two elements combine to form two or more chemical compounds, the masses of one of the elements combine with a fixed mass of the other in a simple ratio.

For example, hydrogen reacts with oxygen to form two compounds: water and hydrogen peroxide.

In this case, the masses of oxygen (i.e. 16g and 32g) that combine with a fixed mass of hydrogen (2g) have a simple ratio, i.e. 16:32 or 1:2.

As we all know, when elements are mixed in different proportions, they form different compounds. For example, when hydrogen is mixed with a different proportion of oxygen, it forms water or hydrogen peroxide.

It demonstrates that there are constituents that combine in a specific manner. These constituents could be atoms. As a result, the law of multiple proportions demonstrates the existence of atoms that combine to form molecules.

Q4. A box contains some identical red coloured balls, labelled as A, each weighing 2 grams. Another box contains identical blue coloured balls, labelled as B, each weighing 5 grams. Consider the combinations AB, AB₂, A₂B and A₂B₃, and show that a law of multiple proportions is applicable.

Answer.

When two elements combine to form two or more compounds, the different masses of one element that combine with a fixed mass of the other bear a simple ratio to one another, according to the law of multiple proportions.

The mass of B when combined with a fixed mass of A (say 1g) is 2.5g, 5g, 1.25g, and 3.75g. They have a 2:4:1 ratio, which is a simple whole-number ratio. Hence. The multiple proportions law is applicable.