Page 12 - Referance Guide For Pharmaceutical Calculations (Naplex, Fpgee and Ptce)

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Milligrams of KCl: The % of triamcinolone in final mixture:

= 80 x 74.5 mg = 100 x 9.54 = 2.1% w/w
= 5960 mg 454
Concentration of KCl in mg/ml:
156.(c) The amount of HCl acid required to prepare
= 5960 mg/250cc 1 gallon, 2.5% HCl is:
= 23.84 mg/cc
= 3840 x 2.5 = 96 gm HCl acid
153.(d) The number of mEq present in 120cc 100
solution (10 mEq/5cc):
The purity of HCl acid is 35% w/w, so to get
= 120 x 10 = 240 mEqs of KCl 96 gm of HCl, we need
5
= 96 x 100 = 274.28 gm of HCl
1 equivalent of KCl = 74.5 gm KCl 35
1 mEq of KCl = 74.5 mg KCl,
The volume of HCl acid needed to prepare 274.28
Milligram of KCl: gm of HCl with a density of 1.25 gm/ml is:

= 240 x 74.5 mg = 274.28 gm = 219.42 ml of HCl
= 17880 mg of KCl 1.25 gm/ml
= 17.88 gm of KCl
157.(d) The % of elementally iron in 325mg of
154.(d) The amount of Povidone in gm in 10% 1 lb ferrous sulfate:
ointment is :
= 65 x 100 = 20% w/w
= 10 x 454 = 45.4 gm of povidone 325
100
158.(d) To solve this problem, we should use the
The cc of povidone required: alligation method,

= 45.4 gm = 22.7cc of Povidone
2 gm/cc

155.(c) The amount of Triamcinolone in 1% of 1 lb
ointment:
To prepare 7.5 (0.5%) parts 0.5 parts (7.5%)
= 454 = 4.54 gm of triamcinolone To prepare 1 gallon ?
100
= 3840 x 0.5 = 256cc 7.5% acetic acid
If we add 5000mg of Triamcinolone: 7.5

= 4.54 gm + 5.0 gm 256cc 7.5% acetic acid solution should mix with
= 9.54 gm of Triamcinolone. (3840-256) 3584cc water to prepare 1 gallon of
0.5% of acetic acid solution.

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