How many carbon atoms are present in 0.35 mole of `C_(6)H_(12)O_(6)`?
`("Given : N"_(A)=6.023xx10^(23))`

Options

A. `6.023 times 10^(23)` carbon atoms
B. `1.26 times 10^(23)` carbon atoms
C. `1.26 times 10^(24)` carbon atoms
D. `6.023 times 10^(24)` carbon atoms

Answer

c

Explanation

`therefore` 1 mol of `C_(6)H_(12)O_(6)` has = 6 N atoms of C
`therefore` 0.35 mol of `C_(6)H_(12)O_(6)` has `=6 times 0.35 N` atoms of C
= 2.1 N atoms `=2.1 times 6.023 times 10^(23)`
`=1.26 times 10^(24)` carbon atoms