How many carbon atoms are present in 0.35 mol of $C_6{H}_{12}{O}_6$?

The correct option is C carbon atoms
$\therefore 1moleof{C}_6{H}_{12}{O}_{6}has=6N_{A}atomsofC\therefore 0.35moleof{C}_6{H}_{1}2O_{6}has=6\times 0.35N_{A}atomsofC=2.1N_{A}atoms=2.1\times 6.022\times {10}^{23}=1.26\times {10}^{24}\text{c}arbonatoms$