Q. The total number of protons in 10 g of calcium carbonate is ($N_{0} = 6.023 \times 10^{23}$)
A) $3.01 \times 10^{24}$
B) $4.06 \times 10^{24}$
C) $2.01 \times 10^{24}$
D) $3.02 \times 10^{24}$
Explanation
(a)
We know that protons in 1 mole $CaCO_{3}$
=atomic number of calcium + atomic number of carbon + 3 (atomic number of oxygen)
$= 20 + 6 + 3(8) = 50$ mol
$\therefore Proton\ in\ 10\ g\ CaCO_{3} = \frac{10 \times 50}{100} \times 6.02 \times 10^{23}$
$= 3.01 \times 10^{24}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.