MCQ
The number of molecules of $CO_2$ present in $44\,g$ of $CO_2$ is
- ✓$6.0 \times {10^{23}}$
- B$3 \times {10^{23}}$
- C$12 \times {10^{23}}$
- D$3 \times {10^{10}}$
mol wt of $C{O_2} = 44$
No. of molecule $ = \frac{{{\rm{wt}}{\rm{. of }}C{O_2}}}{{mol\,wt\,of\,C{O_2}}} \times 6.02 \times {10^{23}}$
$ = \frac{{44}}{{44}} \times 6.02 \times {10^{23}}$$ = 6.02 \times {10^{23}}$
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