## Chemistry: Molecular Approach (4th Edition)

$\underline{\text{0}\text{.95}\ \text{mL}}$.
The weight of the infant is $14\ \text{lb}$.Convert lb into kg as follows: $\left( 14\ \text{lb} \right)\left( \frac{1\ \text{kg}}{2.205\ \text{lb}} \right)=6.3\ kg\,\,\,\left[ \because \,1\text{kg}\,\text{=}\,2.205\ \text{lb}\,\, \right]$ $1\ \text{kg}$ dosage is $15\ \text{mg}$. So, \begin{align} & \text{6}\text{.3}\ kg=\left( \text{6}\text{.3}\ kg \right)\left( \frac{15\ \text{mg}}{1\ \text{kg}} \right) \\ & =95\ mg \end{align} For $\text{80}\ \text{mg}$, the quantity of medicine to be given is $0.80\ \text{mL}$. For $1\ \text{mg}$, the quantity would be $\frac{0.80\ mL}{80\ mg}$. So, for $\text{95}\ \text{mg}$, the volume will be as follows: $\frac{\left( \text{95}\ \text{mg} \right)\left( \text{0}\text{.80}\ \text{mL} \right)}{\text{80}\ \text{mg}}=0.95\ mL$ The volume of the suspension that can be given an infant is $\underline{\text{0}\text{.95}\ \text{mL}}$.