# 3.42g of sucrose are dissolved in

Question:

$3.42 \mathrm{~g}$ 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) $6.022 \times 10^{21}$

Solution:

(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 $=(0.11+1)=1.11 \mathrm{~N}_{\mathrm{A}}$

$=1.11 \times 6.022 \times 10^{23}$

$=6.68 \times 10^{23}$