Question

3.42 g of sucrose are dissolved in 18g 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

The correct option is A $6.68\times {10}^{23}$

Molar mass of sucrose = 342 g

No. of oxygen atoms in 342 g of ($C_{12}{H}_{22}{O}_{11}$) $=11\times N_A$
$\therefore$ No. of oxygen atoms in 3.42 g of ($C_{12}{H}_{22}{O}_{11}$) $=\frac{11\times N_A\times 3.42}{342}$ = $0.11N_A$
No. of oxygen atoms in 18g of $H_{2}O$ $=1N_A$
$\therefore$ Total no. of oxygen atoms in 3.42g of sucrose and 18g of water $=0.11N_A+1N_A=1.11N_A=1.11\times 6.02\times {10}^{23}=6.68\times {10}^{23}$.