(N/A) The molar mass of nalorphene $(C_{19}H_{21}NO_{3})$ is calculated as:
$19 \times 12 + 21 \times 1 + 14 + 3 \times 16 = 311 \, g \, mol^{-1}$
In a $1.5 \times 10^{-3} \, m$ aqueous solution,$1000 \, g$ of water contains $1.5 \times 10^{-3} \, mol$ of nalorphene.
Mass of nalorphene $= 1.5 \times 10^{-3} \, mol \times 311 \, g \, mol^{-1} = 0.4665 \, g$.
Total mass of the solution $= \text{Mass of solvent} + \text{Mass of solute} = 1000 \, g + 0.4665 \, g = 1000.4665 \, g$.
This means $1000.4665 \, g$ of solution contains $0.4665 \, g$ of nalorphene.
To find the mass of solution containing $1.5 \, mg$ $(1.5 \times 10^{-3} \, g)$ of nalorphene:
$\text{Mass of solution} = \frac{1000.4665 \, g \times 1.5 \times 10^{-3} \, g}{0.4665 \, g} \approx 3.22 \, g$.
Thus,the mass of the aqueous solution required is $3.22 \, g$.