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mock exam 1 Questions and Answers Latest updated

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mock exam 1 Questions and Answers Latest updated

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  • February 13, 2024
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  • 2023/2024
  • Exam (elaborations)
  • Questions & answers
  • ATI Teas
  • ATI Teas

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By: Fordenken • 3 months ago

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Year 1 Formulation Mock Exam Calculations and Theory Questions (Q&

1. How many milligrams of mometasone furoate must be added to 50 g of a
0.1% cream to produce a 0.8% mometasone furoate cream?: C1W1 + C2W2 =
C3W3
C1 = 100%, W1 = ?, C2 = 0.1%, W2 = 50g, C3 = 0.8%,
W3 = (50 + W1)

100W1 + 0.1% x 50 = 0.8% x (50 + W1)
100W1 + 5 = 40 + 0.8W1
100W1 - 0.8W1 = 40 - 5
99.2W1 = 35 = 0.35g

0.35g to mg = 350 mg

Ans: 350 mg
2. Will 250 mg of theophylline (solubility 1 in 20) dissolve in 15 ml of water?:
1 in 20
1g = 20ml
0.25g = x
x = 5ml

Ans: Yes, only 5 ml are needed
3. If a drug (MW 325 g/mol) has a solubility of 28 mg/ml in water, is it
possible to prepare a 30 mM solution of this drug in water?: Note;
SOLUBILITY MUST BE GREATER THAN (>) TARGET CONCENTRATION

(28 mg/ 1 ml) = (? mmol / 1000 ml)
x = 28, 000 mg/ 1000 ml

28, g = 86 mmol/1000 ml
Since 86 mM > 30 mM


Then, 30 mM = ? mg/ml
(30mmole/ 1000 ml) = ( x / 1)
x = 0.03 mmoles/1 ml




, Year 1 Formulation Mock Exam Calculations and Theory Questions (Q&

0.03 x 325 = 9.75 mg/ml
Since 28mg/ml > 9.75mg/ml
Therefore, It is possible to prepare such solution.

Answer: Yes, as 28 mg/ml corresponds to 86 mM and 30 mM corresponds to 9.75
mg/m
4. How many milligrams of mometasone furoate must be added to a 0.1%
cream to produce 150 g of a 0.8% mometasone furoate cream?: C1W1 +
C2W2 = C3W3

C1 = 0.1%, W1 = ?, C2 = 100%, W2 = ?, C3 = 0.8%, W3 = 150.

W1 + W2 = 150
W1 = 150 - W2

0.1%x (150 - W2) + 100W2 = 0.8% x 150
15 - 0.1W2 + 100W2 = 120
99.9W2 = 105

W2 = 1.05g to mg = 1050 mg
5. The molar solubility of silver oxalate Ag2(C2O4) in pure water at 25 °C is
2.06 x 10^-4 moles per litre.

Calculate the value of Ksp for Ag2(C2O4) using these data.: Ag2(C2O4) =
[Ag+]^2 [C2O4^2-]
Ag2(C2O4) = 2x ^ 2 * x

Ksp = 2x ^ 2 * x = 4x ^ 3
Ksp = 4 * (2.06 x 10^-4) ^3
Ksp = 3.5 x 10 ^ -11

Ans: 3.5 x 10 ^ -11
6. How much calcium carbonate is present in 120 litres of a 1 in 60,000
solution?: 1 in 60,000; 120 L = 120, 000 ml

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