The number of molecules of `CO_(2)` liberated by complete combustion of 0.1 g of graphite in air is

Options

A. `3.01 xx 10^(22)`
B. `6.02 xx 10^(23)`
C. `6.02 xx 10^(22)`
D. `3.01 xx 10^(23)`

Answer

(c )

Explanation

`C + O_(2) rarr CO_(2)`
1 mole of graphite produces `6.023 xx 10^(23)` molecules of `CO_(2)`
0.1 mole of graphite produces x molecules of `CO_(2)`
` x = (6.023 xx 10^(23) xx 0.1)/1`