$(a)$ Calculate the mass of $0.5$ mole of oxygen atoms.
$(b)$ Calculate the number of molecules of glucose present in its $90 \ g$ (molecular mass of glucose is $180 \ u$). (Avogadro constant $= 6.022 \times 10^{23} \ mol^{-1}$)
$(c)$ Calculate the number of moles of water in $2 \ g$ of water.
(Atomic mass of $H = 1 \ u, O = 16 \ u$)

(N/A) Mass $=$ Molar mass $\times$ Number of moles
$m = M \times n$
$m = 16 \ g/mol \times 0.5 \ mol = 8 \ g$
$(b)$ Number of moles $(n) = \frac{\text{Given mass } (m)}{\text{Molar mass } (M)}$
$n = \frac{90 \ g}{180 \ g/mol} = 0.5 \ mol$
Since $1 \ mol$ contains $6.022 \times 10^{23}$ molecules,
Therefore,$0.5 \ mol$ will contain $0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$ molecules.
$(c)$ Number of moles $(n) = \frac{\text{Given mass } (m)}{\text{Molar mass } (M)}$
Given mass of water $= 2 \ g$,Molar mass of water $(H_2O) = (2 \times 1) + 16 = 18 \ g/mol$
$n = \frac{2}{18} = \frac{1}{9} \approx 0.11 \ mol$