Page No 24: - Chapter 1 Some Basic Concepts Of Chemistry class 11 ncert solutions Chemistry - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 1.22:

If the speed of light is 3.0 ×10⁸ m s^-1, calculate the distance covered by light in 2.00 ns.

Answer:

According to the Question:

Time taken to cover the distance = 2.00 ns

= 2.00 × 10^–9 s

Speed of light = 3.0 × 10⁸ ms^–1

Distance travelled by light in 2.00 ns

= Speed of light × Time taken

= (3.0 × 10⁸ ms ^–1) (2.00 × 10^–9 s)

= 6.00 × 10^–1 m

= 0.600 m

Question 1.23:

In a reaction

A + B₂ → AB₂

Identify the limiting reagent, if any, in the following reaction mixtures.

(i) 300 atoms of A + 200 molecules of B

(ii) 2 mol A + 3 mol B

(iii) 100 atoms of A + 100 molecules of B

(iv) 5 mol A + 2.5 mol B

(v) 2.5 mol A + 5 mol B

Answer:

A limiting reagent determines the extent of a reaction. It is the reactant which is the first to get consumed during a reaction, thereby causing the reaction to stop and limiting the amount of products formed.

(i) According to the given reaction, 1 atom of A reacts with 1 molecule of B. Thus, 200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unused. Hence, B is the limiting reagent.

(ii) According to the reaction, 1 mol of A reacts with 1 mol of B. Thus, 2 mol of A will react with only 2 mol of B. As a result, 1 mol of B will not be consumed. Hence, A is the limiting reagent.

(iii) According to the given reaction, 1 atom of A combines with 1 molecule of B. Thus, all 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric where no limiting reagent is present.

(iv) 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of B will combine with only 2.5 mol of A. As a result, 2.5 mol of A will be left as such. Hence, B is the limiting reagent.

(v) According to the reaction, 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be left as such. Hence, A is the limiting reagent.

Question 1.24:

Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:

N₂₍g₎ + H₂₍g₎ → 2NH₃₍g₎

(i) Calculate the mass of ammonia produced if 2.00 × 10³ g dinitrogen reacts with 1.00 × 10³ g of dihydrogen.

(ii) Will any of the two reactants remain unreacted?

(iii) If yes, which one and what would be its mass?

Answer:

(i) Balancing the given chemical equation,

From the equation, 1 mole (28 g) of dinitrogen reacts with 3 mole (6 g) of dihydrogen to give 2 mole (34 g) of ammonia.

⇒ 2.00 × 10³ g of dinitrogen will react with   dihydrogen i.e.,

2.00 × 10³ g of dinitrogen will react with 428.6 g of dihydrogen.

Given,

Amount of dihydrogen = 1.00 × 10³ g

Hence, N₂ is the limiting reagent.

28 g of N₂ produces 34 g of NH_3.

Hence, mass of ammonia produced by 2000 g of N₂

= 2428.57 g

(ii) N₂ is the limiting reagent and H₂ is the excess reagent. Hence, H₂ will remain unreacted.

(iii) Mass of dihydrogen left unreacted = 1.00 × 10³ g – 428.6 g

= 571.4 g

Question 1.25:

How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?

Answer:

Molar mass of Na₂CO₃ = (2 × 23) + 12.00 + (3× 16)

= 106 g mol^–1

Now, 1 mole of Na₂CO₃ means 106 g of Na₂CO_3.

0.5 mol of Na₂CO₃ 

= 53 g Na₂CO₃

⇒ 0.50 M of Na₂CO₃ = 0.50 mol/L Na₂CO₃

Hence, 0.50 mol of Na₂CO₃ is present in 1 L of water or 53 g of Na₂CO₃ is present in 1 L of water.

Question 1.26:

If ten volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

Answer:

Reaction of dihydrogen with dioxygen can be written as:

Now, two volumes of dihydrogen react with one volume of dioxygen to produce two volumes of water vapour.

Hence, ten volumes of dihydrogen will react with five volumes of dioxygen to produce ten volumes of water vapour.

Question 1.27:

Convert the following into basic units:

(i) 28.7 pm

(ii) 15.15 pm

(iii) 25365 mg

Answer:

(i) 28.7 pm:

1 pm = 10^–12 m

28.7 pm = 28.7 × 10^–12 m

= 2.87 × 10^–11 m

(ii) 15.15 pm:

1 pm = 10^–12 m

15.15 pm = 15.15 × 10^–12 m

= 1.515 × 10^–11 m

(iii) 25365 mg:

1 mg = 10^–3 g

25365 mg = 2.5365 × 10⁴ Ã— 10^–3 g

Since,

1 g = 10^–3 kg

2.5365 × 10¹ g = 2.5365 × 10¹ Ã— 10^–3 kg

25365 mg = 2.5365 × 10^–2 kg

Question 1.28:

Which one of the following will have largest number of atoms?

(i) 1 g Au (s)

(ii) 1 g Na (s)

(iii) 1 g Li (s)

(iv) 1 g of Cl₂(g)

Answer:

1 g of Au (s)  mol of Au (s)

 atoms of Au (s)

= 3.06 × 10²¹atoms of Au (s)

1 g of Na (s) =  mol of Na (s)

 atoms of Na (s)

= 0.262 × 10²³ atoms of Na (s)

= 26.2 × 10²¹ atoms of Na (s)

1 g of Li (s)   mol of Li (s)

 atoms of Li (s)

= 0.86 × 10²³ atoms of Li (s)

= 86.0 × 10²¹ atoms of Li (s)

1 g of Cl₂ (g)   mol of Cl₂ (g)

(Molar mass of Cl₂ molecule = 35.5 × 2 = 71 g mol^–1)

 molecules of Cl₂ (g)

= 0.0848 × 10²³ molecules of Cl₂ (g)

= 8.48 × 10²¹ molecules of Cl₂ (g)

As one molecule of Cl₂ contains two atoms of Cl.

Number of atoms of Cl = 2× 8.48 × 10²¹ =16.96 × 10²¹   atoms of Cl

Hence, 1 g of Li (s) will have the largest number of atoms.

Question 1.29:

Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).

Answer:

Mole fraction of C₂H₅OH 

Number of moles present in 1 L water:

Substituting the value of   in equation (1),

Molarity of solution 

= 2.314 M

Question 1.30:

What will be the mass of one ¹²C atom in g?

Answer:

1 mole of carbon atoms = 6.023 × 10²³ atoms of carbon

= 12 g of carbon

Mass of one ¹²C atom 

= 1.993 × 10^–23 g

Question 1.31:

How many significant figures should be present in the Answer of the following calculations?

(i) 

(ii) 5 × 5.364

(iii) 0.0125 + 0.7864 + 0.0215

Answer:

(i) 

Least precise number of calculation = 0.112

Number of significant figures in the Answer

= Number of significant figures in the least precise number

= 3

(ii) 5 × 5.364

Least precise number of calculation = 5.364

Number of significant figures in the Answer = Number of significant figures in 5.364

= 4

(iii) 0.0125 + 0.7864 + 0.0215

Since the least number of decimal places in each term is four, the number of significant figures in the Answer is also 4.