Calculate the number of the constituent atoms in 53 grams of Na2CO3( At massses of Na= 23u , C=12u, - Science - Atoms and Molecules

Question

Calculate the number of the constituent atoms in 53 grams of
Na2CO3( At massses of Na= 23u , C=12u, O =16u)
I did the questions something like this-
We know that,
n = mass/ molar mass and n= no of particles/ 6 022 * 10 to the
power 23
where n is the no of moles
Now, from above two equations, we can say that, mass/molar
mass=no of particles/6 022 * 10 to the power 23
so, 53/106 = no of particles/6 022 * 10 to the power 23and so, no
of particles = 53/106* 6 022 * 10 to the power 23or, no of
particles = 3 011 * 10 to the power 23Experts please tell me
whether I am right or wrong I f I am wrong, then in which step I
have done wrong Plzzzzz reply fast Thank you

Neha Thukral · Accepted Answer

Molecular mass of
Na2CO3 = 2(23) + 1(12) +3(16) = 46+12+48 =
106g/mol
Given mass of
Na2CO3 =
53g
moles of
Na2CO3 =
53/106 = 0 5mol

Number of constituent atoms:
Number of atoms of Na = 0 5×6
022×1023×2 = 6 022×1023
Number of atoms of C = 0 5×6 022×1023×1 = 3
011×1023
Number of atoms of O = 0 5×6 022×1023×3 = 9
033×1023