Question

3.42 g of sucrose are dissolved in 18 g of water in a beaker. The number of oxygen atoms in the solution are
(a) $6.68\times {10}^{23}$
(b) $6.09\times {10}^{22}$
(c) $6.022\times {10}^{23}$
(d) $6.022\times {10}^{21}$

Answer 1

Number of moles of sucrose $=\frac{Massofsubstance}{Molarmass}$
$=\frac{3.42g}{342gmo{l}^{-1}}=0.01mole$

1 mole of sucrose $\left(C_{12}{H}_{22}{O}_{11}\right)$ contains $=11\times N_A$ atoms~of~oxygen

0.01 mole of sucrose $\left(C_{12}{H}_{22}{O}_{11}\right)$ contains $=0.01\times N_A$ atoms of oxygen
$=0.11\times N_A$ atoms of oxygen

Number of moles of water $=\frac{18g}{18gmo{l}^{-1}}=1mole$

1 mole of water $\left(H_{2}O\right)$ contains $1\times N_A$ atom of oxygen
Total number of oxygen atoms = Number of oxygen atoms from sucrose + Number of oxygen atoms from water
$=0.11N_A+1.0N_A=1.11N_A$

Number of oxygen atoms in solution $=1.11\times$ Avogadro number
$=1.11\times 6.022\times {10}^{23}=6.68\times {10}^{23}$