# Homework 1995/01 - #5 Answers

## Problem 4

t_{1/2} = 11 hr means the kel = 0.06 hr^{-1}
Rearranging (Equation 73) gives

Thus, Dose = 15 x 23 x 0.06 x 6 / 0.9 = 145 mg. Give 150 mg to achieve a
of 15.5 mg/L

Making various assumtions (Equation 71) can be used to calculate Cp_{min} = (150 / 23) x (0.69 / (1 - 0.69)) = 12.8 mg/L

By rough estimation Cp_{max} can be calculated as 18.2 mg/L (15.5 + (15.5 - 12.8))