Question

Total number of protons in $10g$ of calcium
carbonate is $(N_A=6.023\times {10}^{23})$

• A. (a)$12.04\times {10}^{24}$
• B. (b)$4.06\times {10}^{24}$
• C. (c)$2.01\times {10}^{24}$
• D. (d)$3.24\times {10}^{24}$

Answer 1

The correct option is A (a)$12.04\times {10}^{24}$

Molecular mass of CaCO3 = (40+12+16*3)=100 g/mole

No. of moles= 10/100 =0.1 mole

No. Of protons in one molecule of CaCO3=(20+6+8*3)=50

No. of protons in 0.1 mole of CaCO3 will be=(0.1*50)*(6.022*10^23)

=3.011*10^24 protons