Answer:

\textbf{\huge{ANSWER:}}ANSWER:

Formula for Sodium Carbonate: Na_{2}CO_{3}Na

2

CO

3

Molecular mass of Na_{2}CO_{3}Na

2

CO

3

= (23×2) + 12 + (16 × 3)

=》 46 + 12 + 48

=》 106 g

No. of moles in 212g of Na_{2}CO_{3}Na

2

CO

3

:

\begin{gathered}n = \frac{m}{M}\\\end{gathered}

n=

M

m

(n = No. of moles; m = Mass given; M = Molecular mass)

=》 \begin{gathered}n = \frac{212}{106}\\\end{gathered}

n=

106

212

...(1)

Also, we know that:

\begin{gathered}n = \frac{N}{N_{o}}\\\end{gathered}

n=

N

o

N

...(2)

(n = No. of moles; N = No. of particles; N_{o}N

o

= Avagadro's constant,i.e, = 6.023\times 10^{23}6.023×10

23

)

We can keep (1) and (2) equal;

\begin{gathered}\frac{212}{106} = \frac{N}{6.023 \times 10^{23}}\\\end{gathered}

106

212

=

6.023×10

23

N

=》 \begin{gathered}\frac{6.023 \times 10^{23}\times 212}{106}\\\end{gathered}

106

6.023×10

23

×212

=》 \begin{gathered}12.046 \times 10^{23}\\\end{gathered}

12.046×10

23

Now, we know, that there are 2 atoms of Na (Sodium) in Na_{2}CO_{3}Na

2

CO

3

. Therefore;

=》 12.046 \times 10^{23} \times 212.046×10

23

×2

=》 24.092 \times 10^{23}24.092×10

23

Or;

=》\textbf{2.4092 \times 10^{24}}

Thus, there are \textbf{2.4092 \times 10^{24}} Sodium ions in Na_{2}CO_{3}Na

2

CO

3

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