CALCULATE THE NUMBER OF SODIUM IONS THAT ARE PRESENT IN 212 GRAMS OF SODIUM CARBONATE
Answers
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