Question

calculate the gram quantity of sodium carbonate which has same number of atoms as the number of protons present in 10 gram calcium carbonate.

Answer 1

First thing to do is to find out how many molecules of CaCO_3 are in 10g of it.
To do that first find how many moles are there.
The molecular mass of CaCO_3= 100gmol^(-1)
No. of moles, n=m/M=(10g)/100gmol^(-1)=0.1mol

That means there are 0.1*6.022*10^23=0.6022*10^23=6*10^22molecules in 10g.

No. of
"Ca atoms in a molecule"=1
"C atoms in a molecule"=1
"O atoms in a molecule"=3
"Total atoms in a molecule"=5

Now it's to do with protons. In a molecule of CaCO_3 , there are protons.
"Protons of a Ca atom"=20
"Protons of a C atom"=6
"Protons of a O atom"=8
"Total no. of protons in a molecule"=20*1+6*1+8*3=50

So, the total no. of protons in 10g:
"no. of protons"=50*6*10^22=3*10^2*10^22=color(red)(3*10^24

Answer 2

First, let us know that no.of protons = no.of atoms because each atom has one proton and there are no independent protons.

To find out the no. of protons
We need to find the moles of calcium carbonate. It's long work.
The formula  of calcium carbonate is $CaCO _{3}$
Molecular/Molar mass of $CaCO_{3}$ is Ca+C+3O
= 40+12+3(16)
=100

n=Givenmass/Molar mass
n= 10g/100
n=0.1

Now, we know that 0.1 mol of $CaCO_{3}$ will have the same no. of atoms as $Na_{2}CO_{3}$

Molar mass of $Na_{2}CO_{3}$ = 2Na+C+3O = 106

n= given mass/molar mass
n*molar mass = given mass
0.1*106 = given mass
10.6g= given mass.

∴10.6g of $Na_{2}CO_{3}$ will have the same no. of atoms as 10g of $CaCO_{3}$