Total mass of Oxygen atoms

$1$ . Mass of oxygen atoms in $18 \space g$ of glucose:

$\begin{aligned} \quad \quad m_{O1} & = \frac{\text{molar mass of }\ce{O6}}{\text{molar mass of }\ce{C6H12O6}} \times 18 \space g \\ & = \frac{6 \times 16}{6\times 12 + 12 \times 1 + 6 \times 16} \times 18 \space g \\ & = 9.6 \space g \end{aligned}$

$2$ . Mass of oxygen atoms in $90 \space g$ of water:

$\begin{aligned} \quad \quad m_{O2} & = \frac{\text{molar mass of }\ce{O}}{\text{molar mass of }\ce{H2O}} \times 90 \space g \\ & = \frac{16}{2 \times 1 + 16} \times 90 \space g \\ & = 80 \space g \end{aligned}$

Therefore, the total mass of oxygen atoms in glucose solution $= m_{O1} + m_{O2} = \boxed{89.6} \space g$ .

1 mol of C6H12O6 contains 72 grams of C ( 1 mol of carbon is 12 g), 12 grams of Hydrogen( 1 mol of H is 1 g) and 96 grams of Oxygen ( 1 mol of O2 is 16 g). Therefore, 1 mol of glucose = 72+112+96 =180 g. If an aqueous solution contains 18 g of glucose, then the number of grams of Oxygen in this compound is 96/180 X 18g = 9.6 g. In the 90g of water ( H2O), 1 mol of water contains 2 g of H and 16 g of Oxygen or a total of 18 g. The amount of Oxygen in this 90 g of water is 16/18 x 90g = 80 g. Alas, the total MASS OF oxygen ATOMS in this Solution is 9.6+80=89.6g