Answer
(i) 0.028 mol
(ii) $1.7 \times 10^{22} \space molecules$
Work Step by Step
1. Calculate the molar mass of glucose $\big(C_6H_{12}O_6\big)$:
$C: 12.0 \times 6 = 72.0$
$H: 1.0 \times 12 = 12$
$O : 16.0 \times 6 = 96.0$
Molar mass of glucose = $72.0 + 12 + 96.0 =180. \space g/mol$
2. Calculate the amount of $C_6H_{12}O_6$ moles in 5.0 g of glucose:
$$5.0 \space g \times \frac{1 \space mol}{180. \space g} = 0.028 \space mol$$
3. Find the amount of molecules:
Avogadro's constant: $6.022 \times 10^{23}$
$$0.028 \space mol \times \frac{6.022 \times 10^{23} \space molecules}{1 \space mol} = 1.7 \times 10^{22} \space molecules$$
