Question

Calculate the number of atoms of the constituent elements in 53 g of $Na_{2}CO_{3}$.

Answer 1

We have to find the number of moles of 53 g of ${ Na }_{ 2 }{ CO }_{ 3 }$:

Therefore, Molar mass of ${ Na }_{ 2 }{ CO }_{ 3 }\quad =\quad (23\quad \times \quad 2)\quad +\quad 12\quad +\quad (16\quad \times \quad 3)\quad \\=\quad 46\quad +\quad 12\quad +\quad 48\quad \\=\quad 106$

Now, $Moles\quad =\quad \frac { Mass }{ Molarmass }$

               $=\quad \frac { 53 }{ 106 }$

Moles = 0.5

               $1\quad mole\quad =\quad 6.022\quad \times \quad 10^{ 23 }\quad atoms(Avagadro\quad Constant)$

Then, $0.5 moles = 6.022\quad \times \quad 10^{ 23 } \quad \times 0.5 \quad atoms$

               $= \quad 3.011 \quad \times \quad 10^{ 23 } \quad atoms$

Number of moles of 53 g of ${ Na }_{ 2 }{ CO }_{ 3 } \quad = \quad 3.011 \quad \times \quad 10^{ 23 } \quad atoms$.

Answer 2

Explanation:

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