MCQ
The total number of protons in $10\, g$ of calcium carbonate is (${N_0} = 6.023 \times {10^{23}}$)
- A$1.5057 \times {10^{24}}$
- B$2.0478 \times {10^{24}}$
- ✓$3.0115 \times {10^{24}}$
- D$4.0956 \times {10^{24}}$
✓
Answer
Correct option: C.
$3.0115 \times {10^{24}}$
c
(c) $100\,gm$ $CaC{O_3}\; = 6.023 \times {10^{23}}$ molecules
(c) $100\,gm$ $CaC{O_3}\; = 6.023 \times {10^{23}}$ molecules
$\therefore $ $10\,gm$ $CaC{O_3}$=$\frac{{6.023 \times {{10}^{23}}}}{{100}} \times 10$ $ = 6.023 \times {10^{22}}$ molecule
$1$ molecule of $CaC{O_3}$$=50$ Protons $\times 6.023 \times {10^{22}}$ molecule of $CaC{O_3}$$ = 50 \times 6.023 \times {10^{22}}$
$ = 3.0115 \times {10^{24}}$
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