Answer
$10^{23}$
Work Step by Step
The molecular weight of 1 molecule of cellulose is $n X (12[C] + 2 X 1[H] + 16[O])$. Even though we don't know the value of n, we do know that whatever it is contributes to the overall weight of the cellulose molecule. We do know that the carbon contributes $[=12/(12 + 2 + 16) X 100\%]$, or 40%. 40% of 5 g is 2 g, so 2 g of carbon atoms are in the cellulose on the page. So, as the atomic weight of carbon is 12g/mole, and $6 X 10^{23}$ are in a mole, $10^{23}$ carbon atoms are on the page. The equation for this solution would be $ [= (2 g/12 [g/mole]) × 6 × 10^23 (molecules/ mole)]$.
