You would like to prepare 30 mL of an isotonic solution containing 0.45% of drug X. How many mg of sodium chloride should be used to make this isotonic solution? (sodium chloride equivalent (E)-value of drug X = 0.18) 

(3 marks) 

You need to know:

Normal saline 0.9% NaCl is isotonic

Drug X: 30 ml

0.45% = 0.45 g /100ml

0.45 * (30/100) => 0.135 g of drug X in 30 ml

NaCl equivalent of drug X is 0.18

0.135*0.18 = 0.0243 g of NaCl is equivalent to 0.135 g of X

Normal saline 0.9% = 0.9 g/100ml is isotonic

0.9 * (30/100) = 0.27 g NaCl in 30 ml is isotonic

0.27 - 0.0243 = 0.2457 g NaCl is required to make 30 ml solution of 0.45% drug X isotonic.