Answer:
[tex]\large \boxed{\text{90 g}}[/tex]
Explanation:
[tex]\text{M$_{r}$ of H$_{2}$O} = 18.02\\\text{Mass} = \text{5 mol} \times \dfrac{\text{18.02 g}}{\text{1 mol}} = \text{90 g}\\\\\rm \text{There are $\large \boxed{\textbf{90 g}}$ in 5 mol of water.}[/tex]
