How many carbon atoms are present in 0.35 mol of $C_{6} H_{12} O_{6}$ :

6.023 \times 10^{23}

carbon atoms

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1.26 \times 10^{23}

carbon atoms

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1.26 \times 10^{24}

carbon atoms

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6.023 \times 10^{24}

carbon atoms

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Solution

The correct option is B $1.26 \times 10^{23}$ carbon atoms

$∴$ 1 mol of $C_{6} H_{12} O_{6}$ has = 6 $N_{A}$ atoms of C $∴$ 0.35 mol of $C_{6} H_{12} O_{6}$ = $6 \times 0.35$ $N_{A}$ atoms of C = 2.1 $N_{A}$ atoms
= $2.1 \times 6.022 \times 10^{23} = 1.26 \times 10^{24}$ carbon atoms

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