Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 1 - Exercises - Page 40: 103


\[\underline{\text{0}\text{.95}\ \text{mL}}\].

Work Step by Step

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 \[\underline{\text{0}\text{.95}\ \text{mL}}\].
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