$3.42 \mathrm{~g}$ of sucrose are dissolved in $18 \mathrm{~g}$ of water in a beaker. The number of oxygen atoms in the solution are
(a) $6.68 \times 10^{23}$
(b) $6.09 \times 102^{22}$
(c) $6.022 \times 10^{23}$
(d) $6.022 \times 10^{21}$
(a).
No. of moles of sucrose in $3.42 \mathrm{~g}=\frac{(3.42 \mathrm{~g})}{\left(342 \mathrm{gmol}^{-1}\right)}$
$=0.01 \mathrm{~mol}$
No. of oxygen atoms in 1 mole of sucrose
$=0.01 \times 11 \times N_{A}=0.11 \mathrm{~N}_{A}$
No. of oxygen atoms in $0.01$ mole of sucrose
$=0.01 \times 11 \times \mathrm{N}_{\mathrm{A}}=0.11 \mathrm{~N}_{\mathrm{A}}$
No. of moles of $\mathrm{H}_{2} \mathrm{O}$ in $18 \mathrm{~g}=\frac{(18 \mathrm{~g})}{\left(18 \mathrm{~g} \mathrm{~mol}^{-1}\right)}=1 \mathrm{~mol}$.
No. of oxygen atoms in 1 mole of water $=\mathrm{N}_{\mathrm{A}}$
Total no. of oxygen atoms $=(0.11+1)=1.11 \mathrm{~N}_{\mathrm{A}}$
$=1.11 \times 6.022 \times 10^{23}$
$=6.68 \times 10^{23}$
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