Question

How many carbon atoms are present in $0.35\;mol$ of ${C_6}{H_{12}}{O_6}$?
(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 1

For this question we have to know about the mole concept of the molecule. Mole is the fundamental unit of quantity of matter in the International System of Units. The Avogadro constant is the number $6.02214076 \times {10^{23}}$ and it is denoted by the symbol $'{N_A}'$. Atoms, molecules, monatomic or polyatomic ions and other particles can be represented in moles.
$Number\;of\,moles = given\;mole \times \;number\;of\;carbon\,in\,\;1\;mol \times {N_A}\;atoms$
${N_A}$ is the Avogadro constant.

Complete Step By Step Answer:
A mole is the amount of a molecule which contains $6.02214076 \times {10^{23}}$ elementary entities of the given substance.
To solve this question we have to first find the number of moles of individual atoms of the glucose. For that we have to add the masses of each element that makes one glucose molecule. As we know that, in $1\;mol$ of ${C_6}{H_{12}}{O_6}$, number of moles of carbon $C$ is $6$, number of moles of hydrogen $H$ is $12$ and number of moles of oxygen $O$ is $6$.
Now, we will calculate the number of moles of carbon in the given mole of glucose by using the formula,
$Number\;of\,moles = given\;mole \times \;number\;of\;carbon\,in\,\;1\;mol \times {N_A}\;atoms$
$\Rightarrow 0.35 \times 6 \times {N_A}\;atoms$
$\Rightarrow 0.35 \times 6 \times 6.022 \times {10^{23\;}}atoms$
$\Rightarrow 1.26 \times {10^{24\;}}atoms$
Thus, the number of moles of carbon present in $0.35\;mole$ of glucose is $1.26 \times {10^{24}}\;atoms$.
So, option $\left( C \right)$ is correct.

Note:
Mole of a compound always has the same number of entities; never mind what the compound would be. It is found that one mole of any compound has $6.02 \times {10^{23}}$ ions or atoms or molecules etc. This quantity is known as the Avogadro number which is represented by ${N_A}$ in honour of Amedeo Avogadro.