This document provides information and examples of calculations related to pharmaceutical preparations, including: - Alcohol dilution calculations using formulas to determine volumes of stronger alcohols needed. - The alligation method for mixing preparations of different strengths to produce an intermediate strength. - Isotonicity calculations using freezing point depression and molecular concentration methods. - Percentage calculations for weight/volume, volume/volume, etc. - Examples calculating volumes or amounts of ingredients needed for various preparations like saline solutions, alcohol mixtures, and more.

1. Pharmaceutical Calculation
By
Pournima S Shrikhande
Lecturer
Gurunanak technical institute
diploma in pharmacy, Nagpur
Maharashtra, India

2. Alcohol dilution
Alligation method
Calculations for Adjustment to isotonicity
Percentage Calculations
Temperature measurement
Calculation based on density

3. Alcohol dilution
The dilute alcohols are made from 95 % which contains 95 parts by
volume of ethyl alcohol and 5 parts of water

4. Alcohol dilution
The dilute alcohols are made from 95 % which contains 95 parts by volume of
ethyl alcohol and 5 parts of water
Volume of stronger alcohol volume required x percentage required
required / to be used = percentage used

5. Calculate the volume of 95% alcohol required to prepare 400 ml of 50% alcohol.
Solution:
volume required = 400 ml
percentage required = 50
Percentage of alcohol used = 95
By applying the formula :
Volume of stronger alcohol volume required x percentage required
required / to be used = percentage used
= 400x50
95
= 210.52 ml
211 ml of 95% alcohol is diluted with water to produce 400ml . The strength of dilute
alcohol will be 50%

6. Calculate the volume of 95% alcohol required to prepare 500 ml of 40% alcohol
Solution:
volume required = 500 ml
percentage required = 40
Percentage of alcohol used = 95
By applying the formula :
Volume of stronger alcohol volume required x percentage required
required / to be used = percentage used
= 500x40
95
= 210.52 ml
211 ml of 95% alcohol is diluted with water to produce 400ml . The strength of dilute alcohol
will be 50%

7. Alligation method
When the calculation involves mixing of two similar
preparations of different strength, to produce a
preparation of intermediate strength, the alligation
method is used.

8. Calculate the amount of 60%, 50%,40%and 30% alcohol should be mixed
to get 45% alcohol
Using the alligation method
60 15 parts of 60% alcohol
50 05 parts of 50% alcohol
40 05 parts of 40% alcohol
30 15 parts of 30% alcohol
Therefore when 15 parts of 60% alcohol, 05 parts of 50% alcohol, 05 parts
of 40% alcohol, 15 parts of 30% alcohol are mixed together, the resulting
solution will produce 45 % alcohol
45%

9. Calculate the amount of 90%, 80%,30%and 60% alcohol should be mixed
to get 45% alcohol
Using the alligation method
90 15 parts of 90% alcohol
80 15 parts of 80% alcohol
60 35 parts of 60% alcohol
30 45 parts of 30% alcohol
Therefore when 15 parts of 90% alcohol, 15 parts of 80% alcohol, 35 parts
of 60% alcohol, 45 parts of 30% alcohol are mixed together, the resulting
solution will produce 45 % alcohol
45%

10. Calculate the volume of each of 90%, 60%,30%alcohol and water are required
to produce 500ml of 50% alcohol
Using the alligation method
90 50 parts of 90% alcohol
60 20 parts of 60% alcohol
30 10 parts of 30% alcohol
00 40parts of water
120 parts
Therefore when 50 parts of 90% alcohol, 20 parts of 60% alcohol, 10 parts of
30% alcohol and 40 parts of water are mixed together, the resulting solution
will produce 50 % alcohol

11. By applying ratio proportion rule:
Volume of 90 % alcohol required = 120 parts: 500 ml::50 parts: v
v= 500x 50/120
= 25000/120
= 208.33 ml
Volume of 60 % alcohol required = 120 parts: 500 ml::20 parts: v
v= 500x 20/120
= 10000/120
= 83.33 ml

12. Volume of 30 % alcohol required = 120 parts: 500 ml::10 parts: v
v= 500x 10/120
= 5000/120
= 41.67 ml
Volume of water required = 500 ml-(208.33 +83.33 +41.67)
= 166.67 ml
volume of water required is 166.67 ml

13. Calculate the volume of each of 80%, 50%,20%alcohol and water are required
to produce 400ml of 40% alcohol
Using the alligation method
80 40 parts of 80% alcohol
50 20 parts of 50% alcohol
20 10 parts of 20% alcohol
00 40parts of water
___________________
110 parts
Therefore when 40 parts of 80% alcohol, 20 parts of 50% alcohol, 10 parts of 20% alcohol
and 40 parts of water are mixed together, the resulting solution will produce 40 % alcohol

14. By applying ratio proportion rule:
Volume of 80 % alcohol required = 110 parts: 400 ml::40 parts: v
∴ v= 400x 40/110
= 16000/110
=145.45 ml
Volume of 50 % alcohol required = 110 parts: 400 ml::20 parts: v
∴ v= 400x 20/110
= 8000/110
= 72.72ml

15. Volume of 20 % alcohol required = 110 parts: 400 ml::10 parts: v
∴ v= 400x 10/110
= 4000/110
= 36.36 ml
Volume of water required = 400 ml-(145.45 +72.72 +36.36)
= 400- 254.53 ml
= 145.47 ml
volume of water required is 145 ml

16. Calculate the volume of each of 65%, 95%,35%alcohol and water are required
to produce 600ml of 45% alcohol
Practice problem

17. Calculate the volume of each of 65%, 95%,35%alcohol and water are required
to produce 600ml of 45% alcohol
Using the alligation method
95 45 parts of 95% alcohol
65 10 parts of 65% alcohol
35 20 parts of 35% alcohol
00 50parts of water
___________________
125 parts
Therefore when 45 parts of 95% alcohol, 10 parts of 65% alcohol, 20 parts of 35% alcohol
and 50 parts of water are mixed together, the resulting solution will produce 40% alcohol

18. By applying ratio proportion rule:
Volume of 95 % alcohol required = 125 parts: 600 ml::45 parts: v
∴ v= 600x 45/125
= 16000/110
=216 ml
Volume of 65 % alcohol required = 125parts: 600 ml::10 parts: v
∴ v= 600x 10/125
= 6000/125
= 48ml

19. Volume of 35 % alcohol required = 125 parts: 600 ml::20 parts: v
∴ v= 600x 20/125
= 12000/125
= 96 ml
Volume of water required = 600 ml-(216+48+96)
= 600- 360 ml
= 240 ml
Volume of water required is 240 ml

20. When more than two ingredients with known strengths are available and
mixture possessing the strength lying between the given strengths is
required the following method is used as under
Steps:
1) Write percentage strengths in descending order on the left hand side of
vertical line and the required percentage between the two vertical lines.
2) Then make the pairs of strength in such a way that in a pair one strength
must be lower and other strength must be higher than required strength .

21. In what proportions should 15% 8%and 3% alcohol be mixed to obtained 6%
alcohol ?
Thus the proportion for mixing would be 3:3:11 to get 6% alcohol by mixing
15% 8%and 3% alcohol respectively
15% 3 parts
8% 6% 3 parts
3% 9+2= 11 parts

22. A pharmacist has 3 lots of ointment containing 50% 30% 20% of an
active drug respectively in what proportion these are mixed to obtain
an ointment containing 25% of active drug.
50 05 part for 20%
30 05 part for 30%
20 25+ 5= 30 part for 20%
Thus the proportion for mixing would be 5:5:30 to get 25% of active
drug by mixing 50% 30%and 20% of an active drug respectively
25

23. In what proportions should 90% 60%and 30% alcohol be mixed
to obtained 55% alcohol ?
Practice problem

24. In what proportions should 90% 60%and 30% alcohol be mixed to obtained
55% alcohol ?
90 25 part for 90%
60 25 part for 60%
30 35+ 5= 40 part for 30%
Thus the proportion for mixing would be 25:25:40 to get 55% of alcohol
by mixing 90% 60%and 30% of an alcohol respectively
55

25. Isotonic solutions
All the ophthalmic solutions an injectable should be isotonic.
Solutions having the same osmotic pressure are called as
iso-osmotic
osmotic pressure is defined to be the minimum pressure
required to maintain an equilibrium with no net movement of
solvent.

26. • Isotonic/ isosmotic solution :
Two solutions having the same osmotic pressure are called Isotonic or
isosmotic.
• Paratonic solution :
The solutions which do not have the same osmotic pressure are called
as Paratonic.

27. • Hypertonic solution:
Hypertonic solution contain more quantity of solute than required for
making them isotonic
• Hypotonic solution:
• Hypotonic solution contain less quantity of solute than required for
making them isotonic

28. Principle for adjustment to isotonincity
• Solution for IV injection: isotonicity is always desirable
• Solution for SC injection: not essential
• Since they are injected into fatty tissues and not in blood stream
• Solution for IM injection: the aqueous solution must be slightly
hypertonic to promote rapid absorption

29. • Solution for Intra- cutaneous injection:
• The parenteral preparation which are meant for diagnostic purpose
should be isotonic in order to avoid the false reaction.
• Solution for intra-thecal injection: must be isotonic
• Small volume of paratonic solution will disturb the osmotic pressure
and may cause vomiting and other side effects.

30. • Solution used for nasal drops: isotonic. paratonic solution may cause
irritation
• Solution used as eye lotion: isotonic.
• Solution used as eye drops: may not be isotonic
• As small volume is used which quickly get diluted by the lachrymal
secretion.

31. Calculation for adjustment of isotonicity
Based on freezing point method:
Percentage w/v of adjusting substance required= 0.52- a
b
a = freezing point of the unadjusted solution
b= freezing point of a 1% w/v solution of the adjusting substance

32. Find the concentration of sodium chloride required to make 1 %
solution of boric acid, iso-osmotic with blood plasma.
(Given: The freezing point of1% w/v solution of boric acid is 0.288 o C)
The freezing point of1% w/v solution of sodium chloride is 0.576 o C)
Calculation:
By applying the formula:
Percentage w/v sodium chloride required= 0.52- a
b
= 0.52- 0.288
0.576
=0.402 %w/v

33. Find the concentration of sodium chloride required to make 1.5 % solution of
cocaine hydrochloride , iso-osmotic with blood plasma.
(Given: The freezing point of1% w/v solution of cocaine hydrochloride is -
0.09o C)
The freezing point of1% w/v solution of sodium chloride is -0.576o C)
Calculation:
By applying the formula:
Percentage w/v of sodium chloride required= 0.52- a
b
= 0.52- (0.09x1.5)
0.576
= 0.668 %w/v

34. Find the concentration of sodium chloride required to make 50 ml of
isotonic solution containing 0.5 % ephedrine hydrochloride and 0.5%
chlorobutol.
(Given: The freezing point of1% w/v solution of ephedrine
hydrochloride is -0.165 o C)
The freezing point of1% w/v solution of chlorobutol is -0.138 o C)
Calculation:
By applying the formula:
Percentage w/v of sodium chloride required= 0.52- a
b

35. Calculation:
As the concentration of ephedrine hydrochloride in the preparation 0.5% w/v,
The depression in the freezing point of ephedrine hydrochloride = 0.165x0.5
=0.0825 o C
As the concentration of chlorobutol in the preparation 0.5% w/v,
The depression in the freezing point of chlorobutol = 0.138x 0.5
=0.069 o C
∴ total depression in the freezing point of the substances
will be = 0.0825+ 0.069
= 0.1515

36. By applying the formula:
Percentage w/v of sodium chloride required= 0.52- a
b
= 0.52- 0.1515
0.576
= 0.3685
0.576
= 0.6397 %w/v

37. • Weight of sodium chloride required to make 100 ml of solution=0.6397g
• Weight of sodium chloride required to make 50 ml of solution=0.6397
2
= 0.3198g

38. Find the concentration of sodium chloride required to make 1 % solution
of adrenaline, iso-osmotic with blood plasma.
(Given : The freezing point of a 1% w/v solution of adrenaline is 0.098 o C)
Practice problem

39. Calculation:
By applying the formula:
Percentage w/v of sodium chloride required = 0.52- a
b
= 0.52-0.098
0.576
= 0.7326%w/v
0.7326%w/v of sodium chloride required to make 1 % solution of
adrenaline, iso-osmotic with blood plasma

40. Based on molecular concentration
Percentage w/v of adjusting substance required= 0.03M
N
Where,
M= gram molecular weight of substance
N= number of ions in which the substance is ionised

41. Find the concentration of sodium chloride required to produce a solution iso-
osmotic with blood plasma.
Given :molecular weight of sodium chloride = 58.5
sodium chloride is ionizing substance and it gets dissociates into 2 ions
Calculation :
Hence the formula used is W= 0.03 M
N
W= 0.03 x 58.5
2
W= 0.88gm/100ml

42. Find the proportion of dextrose needed to form a solution iso-osmotic with
blood plasma.
Given: molecular weight of dextrose= 180
Calculation :
Dextrose is non ionizing substance.
Hence the formula used is W= 0.03 M
W= 0.03 x 180
W= 5.4 gm/100ml

43. Percentage calculations
• Percent w/v (Percentage weight in volume)
• Percent v/v (Percentage volume in volume)
• Percent v/w (Percentage volume in weight)
• Percent w/w (Percentage weight in weight)

44. formulae
Preparation of percentage solution by diluting the concentrated solution
Strength of dilute solution = Strength of concentrate
Degree of dilution

45. Volume of stronger solution/ = Volume required x Percentage required
alcohol to be used Percentage used
Weight of stronger acid to be used = Weight required x Percentage required
Percentage used

46. Calculate the quantity of sodium chloride required to prepare 400 ml 0f 0.9%
solution.
Calculation
0.9 % solution means 0.9 gm in 100 ml solvent
amount of sodium chloride required to produce 400 ml
solution= 0.9x400
100
= 3.6 g
Hence, 3.6 g of sodium chloride is dissolved in water to produce 400 ml
makes 0.9 % w/v solution

47. Prepare 400 ml of 5% solution and label with a direction for preparing 2 litre
quantity of a 1 in 2000 solution.
Calculation:
Strength of concentrated solution = 5%
Strength of dilute solution = 1 in 2000
= 0.005
=0.005 x100
=0.05%
Volume of dilute solution required = 2 liter= 2000ml
Volume of concentrated solution required =?

48. By applying formula,
volume of stronger solution
required = volume of dilute solution x percentage required
percentage used
= 2000x0.05
5
= 20 ml
Therefore dilute solution is obtained by diluting 20 ml of concentrated
solution to 2 litre

49. Prepare 1 lit. solution 1 in 4000 using 0.1% w/v solution.
Calculation
strength of concentrate = 0.1%
strength of dilute solution 1 in 4000 = 0.00025
=0.00025 x100
= 0.025%
Volume of dilute solution required = 1 liter= 1000ml
Volume of concentrated solution required =?

50. By applying formula,
volume of stronger
solution required = volume of dilute solution x percentage required
percentage used
= 1000x0.025
0.1
= 250 ml
Therefore dilute solution is obtained by diluting 250 ml of concentrated
solution to 1litre

51. How much 5% solution is required to prepare 600 ml of a1in 800 solution?
Calculation:
Data given:
Strength of concentrated solution: 5%
Strength of dilute solution: 1 in 800
= 0.00125x100
=0.125%
Volume of dilute solution required = 600ml
Volume of concentrated solution required =?

52. By applying formula,
volume of stronger
solution required = volume of dilute solution x percentage required
percentage used
= 600x0.125
5
= 15 ml
Therefore 15 ml of 5% solution is diluted to 600 ml to prepare 1:800 solution

53. Prepare 200ml of 5% solution of chloroform in 95% alcohol
Calculation:
5% v/v solution = 5 ml of chloroform dissolved in 100 ml of alcohol
∴ solution required to prepared 5%of 200ml solution = 5x2
= 10ml
Hence, dissolve 10 ml of chloroform in sufficient quantity of 95% alcohol to
make200ml of a 5% v/v solution

54. Temperature measurement :
9 (o C) = 5 (o F) -160
(o F) is the no of degree Fahrenheit , (o C) is the no of degree Centigrade
Convert 120 o F into o C
9 (o C) = 5 (o F) -160
= 5 (120)- 160
= 600 – 160
= 440
o C = 440/ 9
o C = 48.9 o C

55. Convert 30 o C into o F
9 (o C) = 5 (o F) -160
9x30 = 5 (o F) -160
270 + 160 = 5 (o F)
270 + 160/5 = o F
430/ 5 = o F
86 o F

56. Calculation based on density
• Density is defined as the mass of a substance per unit volume.
• Specific gravity is defined as the ratio of the mass of a substance in air to
that of equal volume of water

57. Weight
Density = ________
Volume
Weight = Density x Volume
Volume= Weight
________
Density

58. Calculate the volume of 2kg of glycerine. The density of glycerine is 1. 25
g/ml.
Weight
Volume = ------------------------
Density
= 2000/ 1.25
= 1600 ml

59. Calculate the weight of 200 ml of alcohol whose density is
0.816 g/ml.
Ans:
Weight = Density x volume
= 0.816x 200
= 163.2 g

60. Thank you