Question

How many milligrams of dextrose monohydrate should be used in compounding the following prescription? Round vour answer to the nearest whole number. Rx. Ephedrine hydrochloride (E 0.29) Dextrose, monohydrate (E 0.16) Purified water Make isotonic solution 0.5 g qs ad 30 mL

Answer 1

The structures of ephedrine hydrochloride and dextrose monohydrate along with the molecular formula and molecular weights of the compounds are given below

dextrose monohydrate ephedrine hydrochloride ОН ОН Н НО NHCH3 НО о CHS . НС НО НО H2о C6H1407 MW 198.17 g/mol MF = MF C10H16CINO MW 201.69 g/mol

Isotonic solution means the osmotic pressures of dextrose monohydrate and ephedrine hydrochloride should be same.

Lets consider the osmotic pressure of dextrose monohydrate is

ID= iMn RT

i = Van't Hoff factor = 5

M_D = concentration in molarity = n/V; n = number of moles of dextrose monohydrate, V = volume =30 mL

R = gas constant = 0.082 L atm mol^-1 K^-1

T = temperature in K

Thus,

5n RT Пр 30

Lets consider the osmotic pressure of ephedrine hydrohloride is

II=iMERT

i = Van't Hoff factor = 1

M_E = concentration in molarity = n/V; n = number of moles of ephedrine hydrochloride = 0.5 g/201.69 g mol^-1 = 2.48$\times$10^-3 moles,

V = volume =30 mL

Thus,

1x2.48 х 10 -3RT Пр 30

Now, the condition is both the osmotic pressures are same. Therefore,

5n RT 1 x 2.48 x 10-3RT 30 30

2.48 x 10 0.496 x 10mol 5

Thus, the weight of dextrose monohydrate should be = MW of dextrose monohydrate$\times$ number of moles of dextrose monohydrate

=

198.17gmollx0.496x10-3mol 98.2x10g 0.098g 0.10g(answer)