Question

Calculate the gm quantity of Na2Co3 which has same
number of atoms as the number of protons present in
10 gm CaCO3
(A) 20 gm
(B) 88.33 gm
(C) 44 gm
(D) 60 gm
​

Answer 1

Answer:

First we will find the no. of moles in 10 grams of CaCO 3

Molecular mass of CaCO 3 = (40+12+16*3)=100 g/mole

No. of moles= 10/100 =0.1 mole

No. Of protons in one molecule of CaCO 3 =

(20+6+8*3)=50

No. of protons in 0.1 mole of CaCO 3 will be

=0.1*50*6.022*10 23

=3.011*10 24 protons

Answer 2

Concept:

The number of moles is calculated by the formula, No of moles = Mass/ molecular mass. The number of x atoms in x gm of Na₂Co₃ is equivalent to the number of protons in 10 gm of CaCO₃.

Given:

Weight of Calcium carbonate = 10gm

Find:

We need to determine the gm quantity of Na₂Co₃ which has the same number of atoms as the number of protons present in 10 gm CaCO₃.

Solution:

Let x g be the amount of Na₂Co₃.

The number of x atoms in x gm of Na₂CO₃ is equivalent to the number of protons in 10 gm of CaCO₃

Therefore, the number of atoms in x gm of Na₂Co₃ = x/106 × Na × 6

the number of atoms in x gm of Na₂CO₃ = 6x/106 × Na

Number of moles = Mass of the compound / Molecular mass of that compound

Thus, 1 molecule of Na₂CO₃ contains about 6 atoms of Na₂CO₃ and the Number of molecules = No. of moles × Avogadro's Number

While the number of protons in 1 molecule of CaCO₃ = 20 + 6 + 8(3)

= 20 + 6 + 24 = 50

Therefore, the number of protons in 10g of CaCO₃ = 10/100 × Na × 50

Therefore, 6x/106 × Na = 10/100 × Na × 50

Therefore, x = 5 × 106 /6

x = 88.33 g

Thus, the amount of Na₂CO₃ is 88.33 g.

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