Question

If Avogadro number is $6\times {10}^{23}$, then the number of protons, neutrons and electrons in $14$ g of ${{}_{6}C}^{14}$ is, respectively:

• A. $36\times {10}^{23}$, $48\times {10}^{23}$, $36\times {10}^{23}$
• B. $36\times {10}^{23}$, $36\times {10}^{23}$, $36\times {10}^{23}$
• C. $48\times {10}^{23}$, $36\times {10}^{23}$, $48\times {10}^{23}$
• D. $48\times {10}^{23}$, $48\times {10}^{23}$, $36\times {10}^{23}$

Answer 1

Answer: (A)

$36\times {10}^{23}$, $48\times {10}^{23}$, $36\times {10}^{23}$

14 g of ${}_6^{14}C$ correpsonds to 1 mole which contains avogadro's number of molecules. It contains 6 moles of protons, $14-6=8$ moles of neutrons and 6 moles of electrons.
Number of protons $=6\times 6\times {10}^{23}=36\times {10}^{23}$
Number of neutrons $=8\times 6\times {10}^{23}=48\times {10}^{23}$
Number of electrons $=6\times 6\times {10}^{23}=36\times {10}^{23}$