Question

Calculate the no. of constituient atoms in 53 g of na2co3

Answer 1

Answer :

• No. of atoms = 3.011 × 10²³.

Step-by-step explanation :

Given that,

• Mass of $\rm{Na_2CO_3}$ = 53 g.

Also,

• Avogadro's number = $\rm{6.022\times{}10^{23}\:mol^{-1}}$

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Molar mass of $\rm{Na_2CO_3}$ =

⇒ 23 × 2 + 12 + 16 × 3

⇒ 106 g/mol.

Now, $\rm{No.\:of\:moles=\dfrac{Mass}{Molar\:mass}}$

Putting the values, we get,

No. of moles = $\rm{\dfrac{53\:g}{106\:g\:mol^{-1}}=0.5\:mol}$

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Again, we know 1 mol of $\rm{Na_2CO_3}$ contains 6.022 × 10²³ atoms

∴ 0.5 moles will contain =

⇒ 0.5 × 6.022 × 10²³ atoms

⇒ 3.011 × 10²³ atoms.

Answer 2

Answer:

Avagadro's constant states that 1 mole of any substance has 6.022 × 10²³ molecules/atoms.

Find the number of moles of 53 g of Na2CO3:

molar mass Na2CO3 = 23 × 2  + 12 + 16 × 3 = (46+12+48) = 106

moles = mass/molar mass

  

           = 53/106

           = 0.5 moles

1 mole = 6.022 ×10²³ atoms

Then 0.5 moles = 6.022 × 10²³ × 0.5 atoms

                     

                          = 3.011 × 10²³ atoms of Na2CO3