Chemistry Basics

NCERT Solution 4

Question 28: Which one of the following will have the largest number of atoms?

1 g Au (s)
1 g Na (s)
1 g Li (s)
1 g of Cl₂ (g)

Answer: Atomic masses: Au = 197, Na = 23, Li = 7, Cl = 35.5 u

We know number of atoms in mass equivalent to atomic mass is a constant, i.e. 6.022 × 10²³

For obtaining no. of atoms in 1 g of a particular atom, we need to divide 1 mole by atomic mass.

Smaller denominator will give a larger number compared to a larger denominator. Here, atomic mass of Li is the smallest and hence, 1 g of Lithium will contain the largest number of atoms.

Question 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: `x_e=(n_e)/(n_e+n_w)=0.040` given

Here, e represents ethanol and w represents water

As the solution is dilute so 1 L of solution is composed of almost all of water

No. of moles in 1 L of water `=(1000g)/(18g\text(mol)^(-1))=55.55` mole

Substituting this value in first equation, we get:

`(n_e)/(n_e+55.55)=0.040`

Or, `n_e=0.040(n_e+55.55)`

Or, `n_e=0.040n_e+2.222`

Or, `0.96n_e=2.222`

Or, n_e = 2.222 × 0.96 = 2.31 mol

So, molarity of the solution = 2.31 M

Question 30: What will be the mass of one 12C atom in g?

Answer: Atomic mass of carbon is 12 g which contains 1 mole of carbon atom

It means 12 g contains 6.022 × 10²³ carbon atoms

So, mass of 1 carbon atom

= 12 ÷ 6.022 × 10²³

= 1.993 × 10^-23

Question 31: How many significant figures should be present in the answer of the following calculations?

`(0.02856 × 298.15 × 0.112)/(0.5785)`
5 × 5.364
0.0125 + 0.7864 + 0.0215

Answer: (i) 2, (ii) 4, (iii) 4

Question 32: Use the data given in the following table to calculate the molar mass of naturally occuring argon isotopes:

Isotope	Isotopic molar mass	Abundance
³⁶Ar	35.96755 g mol^–1	0.337%
³⁸Ar	37.96272 g mol^–1	0.063%
⁴⁰Ar	39.9624 g mol^–1	99.600%

Answer: It will be sum of following:

35.96755 × 0.337% = 0.121

37.96272 × 0.063% = 0.024

39.9624 × 99.600% = 39.803

Sum = 39.803 + 0.121 + 0.024 = 39.948 g mol^-1

Question 33: Calculate the number of atoms in each of the following (i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He.

Answer: Number of atoms in 52 moles of Ar = 52 × 6.022 × 10²³

= 3.131 × 10²⁵

Atomic mass of helium = 4 u

So, 52 u of helium = 52 ÷ 4 = 13 atoms

No. of atoms in 52 g helium = (52 ÷ 4) × 6.022 × 10²³

= 7.83 × 10²⁵

Question 34: A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g.

Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.

Answer: Molar mass of CO₂ = 44 g and that of H₂O = 18 g

44 g of carbon dioxide contains 12 g carbon

So, 3.38 g carbon dioxide contains `(12)/(44)×3.38=0.9218` g carbon

Similarly, 18 g water contains 2 g hydrogen

So, 0.690 g water contains `2/(18)×0.690=0.0767` g hydrogen

Since welding fuel is composed of only carbon and hydrogen so the mass of welding fuel

= 0.9218 + 0.0767 = 0.9985 g

No. of moles of carbon in this compound = 0.9218 ÷ 12 = 0.768

No. of moles of hydrogen in this compound = 0.0767 ÷ 1 = 0.767

Ratio of moles of carbon and hydrogen = 1 : 1

So, empirical formula of compound is CH

Molar mass of gas can be calculated as follows:

10 L of gas weighs 11.6 g

So 22.4 L (volume of standard gas) weighs

11.6 ÷ 10 × 22.4 = 25.98 g = 26 g (approx)

Molecular Formula can be calculated as follows:

Empirical formula CH means mass is 12 + 1 = 13 g

But we have calculated mass as 26 g (earlier)

So, molecular formula is C₂H₂

Question 35: Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction,

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

What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl?

Answer: Molar mass of HCl = 36.5 g

Molar Mass of CaCO₃ = 100 g

So, 0.75 M HCl contains 36.5 × 0.75 = 27.375 g HCl (in 1 L water)

So, 25 mL water contains 27.375 ÷ 40 = 0.684 g HCl

As per equation, 2 mole of HCl reacts with 1 mole of calcium carbonate

This means 73 g HCl reacts with 100 g calcium carbonate

So, 0.684 g HCl reacts with `(100)/(73)×0.684=0.937` g calcium carbonate

Question 36: Chlorine is prepared in the laboratory by treating manganese dioxide (MnO₂) with aqueous hydrochloric acid according to the reaction

4 HCl (aq) + MnO₂ (s) → 2H₂O (l) + MnCl₂ (aq) + Cl₂ (g)

How many grams of HCl react with 5.0 g of manganese dioxide?

Answer: Molar mass of MnO₂ = 55 + 2 × 16 = 55 + 32 =87 g

Molar mass of HCl = 36.5 g

4 moles of HCl = 4 × 36.5 = 146 g

87 g of manganese dioxide reacts with 146 g HCl

So, 5 g of manganese dioxide reacts with `(146)/(87)×5=8.40` g HCl