Question

How many carbon atoms are present in $0.35 \, mole$ of $ \, C_{6}H_{12}O_{6}?$
(Given: $\text{N}_{\text{A}} = 6.023 \times 10^{23}$ )

• A. $1.26 \, \times 10^{2} \, carbon \, atoms$
• B. $1.26 \, \times 10^{24} \, carbon \, atoms$
• C. $1.26 \, \times 10^{44} \, carbon \, atoms$
• D. $1.26 \, \times 10^{48} \, carbon \, atoms$

Answer 1

Answer: (B)

$\because \, 1moleofC_{6}H_{12}O_{6}has= \, 6 \, $
$0.35 \, mole \, of \, C_{6}H_{12}O_{6} \, has \, = \, 6 \, \times \, 0.35 \, $
$= \, 2.1 \, \times \, 6.023 \, \times 10^{23}= \, 1.26 \, \times 10^{24} \, carbon \, atoms$