Question

How many carbon atoms are present in 0.35 mol of c6h12o6 how to calculate?

Answer 1

Answer:

The carbon atoms in the glucose is 6 so the carbon atoms in one mole of glucose will be 6x 6.022x10^23 atoms where we know that for any moles of atom we have 6.022*10^23 atoms of that molecules.

So, as in the given question 0.35 mole is asked hence number of carbon atoms in 0.35 moles of glucose will be = 0.35x 6x6.022x10^23  

Which will approximately be equal to  12.64 x 10^23 atoms.

Answer 2

Answer:

$\bold{1.26 \times 10^{23}}$ number of carbon atoms are present.

Explanation:

One mole of glucose ($\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$) has six N atoms of glucose which means six carbon atom/molecule of glucose. Number of atoms per molecule in a mole representing the Avogadro’s constant ($6.02 \times 10^{23}$).

Therefore 0.35 mole of glucose will have = n x L x N (n= number of mole, L= Avogadro’s constant and N= number of atom.

$=0.35 \times 6.02 \times 10^{23} \times 6=1.26 \times 10^{23}$