CALCULATIONS

LAWS
Vd = Dose/Co
Vd = Volume of distribution
Co = Conc. Of drug in plasma at zero time
Loading dose= Vd * Css
Loading dose= Vd [C2-C1]
Loading dose is the dose needed to reach steady state
• Css = Concentration of the drug in blood at steady state
• C1= Concentration of the drug in plasma
• C2= Concentration of the drug needed to add to C1 to reach required conc.
· At steady state rate of drug input= rate of elimination

· Time required to reach steady state (T ss)= 4.5 or 5 t1/2

· Half life (T1/2)= the time required for the concentration
of a substance in the body to decrease by half
Therapeutic index(TI)= LD50/ED50

LD50=Median Lethal Dose is the amount of an

agent that is sufficient to kill 50 percent of a
population of animals
ED50= Median Effective Dose is the dose that
produces a quintal effect in 50% of the
population

Drug with narrow TI = highly dangerous

Bioavailability= AUC/ Concentration
Bioavailability=AUC (oral) / AUC (iv) * 100

Bioavailability = Area Under Curve =

Plasma conc of drug by any route /Plasma conc of drug by iv AUC
Specific gravity= Wt. of substance (kg) /wt of equal amount of water(L)
Specific gravity= Mass unit volume of sub/mass unit volume of water
Specific gravity= Density of sub/ Density of equal amount of water

Density= Mass (gm) / volume (ml) or Kg/L

MEq= Wt. (mg) * valency / M. wt
MEq= Mill equivalent
 CLEARANCE LAWS
· Clearance (Cls)= 0.693 * Vd / t1/2
· Vd= Dose/Co
· Cls= Rate of elimination/ drug conc
· Cls= Renal Cls+ Non renal Cls
· Cls= Ke * Vd [Ke= elimination rate constant]
· Cr clearance for male= (140-age) * weight /72 * Sr. Cr
· Cr. Cls for female= Cr. Cls for male * 0.85
 MOLALITY LAWS
· Molarity (m)=No. of moles of solute/ L volume of solution
· Molality (M)= No. of moles of solute/ Kg of solvent

· Mosm= Mill moles * No. of species

· Examples of no. of species:
Ex1: NaCl = 1Na + Cl= 2----- Ex2: CaCl2 = 1 Ca + 2Cl= 3
Check problems No.11 & 12
Some conversations:
Weight:
· Kg= 2.2 pound (lb)
· Grain= 0.065gm
Volume:
· Tea spoonful (tsp)= 5ml
· Table spoonful (tbsp)= 15ml
· 16 drop (dp)= 1ml
· 1 fluid ounce =30ml
· 1L= 0.22 Gallon
· 1L= 10 Deciliter
Tempreture:
· 5F= 9C + 160
· F= Fahrenheit
· C= celsius

Length:
· 1FOOT (FT)= 12 inch
· 1 inch= 2.54 cm
Others:
· PPM= Part Per Million= mg / L
· 10% w/w= 10 gm in 90 gm (total wt. = 100 gm) w/w= gm in gm
· 10% w/v = 10 gm in 100 ml (total volume= 100 ml) w/v= gm in ml
· 10% v/v= 10 ml in 90 ml (total volume = 100 ml) v/v= ml in ml
Some molecular weights you may use:
· HCl= 36.4
· NaCl= 58.5
· CaCl2= 111
· KCl= 74.5
· NH4Cl= 53.5
· MgCl2= 95.2
Some other laws haven’t been used till now but may be
useful for you
· Child dose= wt (lb) / 150* adult dose
· Child dose= age / (age+ 12) * adult dose
· E= Extraction ratio= drug elimination of an organ (e.g. Liver)
· E= [Arterial drug conc – Venous drug conc] / Arterial drug conc
· Cls of liver= E x hepatic blood flow
PROBLEMS
1- Amount of drug is 5 mg in 1ml what the amount of drug
in 1 tsp in micrograms.
A. 25mcg
B. 500mcg
C. 2000mcg
D. 25000mcg

Answer:
1 tsp= 5ml
5mg in 1ml
Xmg in 5ml
X= 5* 5/1= 25mg
25*1000 = 25000mcg
2- A solution is made by dissolving 17.52g of
NaCl exactly 2000ml. What is the Molarity of
the solution?

A. 3.33
B. 0.15
C. 1.60
D. 1.6*10-4
Answer:
Molarity= Mole/ Volume (L)

1 Mole= Molecular weight of subs. In 1 grams

No.of Moles= Wt / M.wt

Molecular weight of NaCl = 58.5

Mole= 17.52/58.5= 0.299

Molarity= 0.299/2= 0.149

3- 5ml of injection that concentration 0.4%
calculate the amount of drug?

A. 0.2mg
B. 2mg
C. 200mg
D. 20mg
Answer:
0.4 g in 100 ml
X g in 5ml
X= 5* 0.4/100= 0.02 gm
0.02*1000 = 20mg
 4. An elixir contains 0.1 mg of drug per ml. How many
micrograms are there in one tsp of the elixir?
A. 0.0005 mcg
B. 0.5 mcg
C. 500 mcg
D. 5 mcg
Answer:
0.1 mg in 1 ml
X mg in 5ml
X= 0.1 * 5/1= 0.5 mg = 0.5*1000 = 500 mcg
5- Solution contains D50W another one contain D5W we
want to prepare solution contain D15W its volume is
450ml. How much ml we need of each solution

A. D50W/D5W= 10/35
B. D50W/D5W= 20/25
C. D50W/D5W= 5/40
D. D50W/D5W= 10/40
C1=50 C2= 5
V1= X V2= Y
C = 15
V =450 ml
Answer by equation method:

Try the choices ratio in the equation:

(C1*V1)+(C2*V2)= (C*V)
(50 * 10) + (5 * 35)= (15* 45)
675 = 675
(50 *20 ) + (5 * 25)= (15* 45)
1125= 675
 Answer by Allegation method:

50 5
15
10 + 35 = 45
X + Y = 450
X= 10*450/ 45= 100
Y= 35*450/45= 350
6- Prescription Hydrocortisone 2% Cold cream 60gm.
You have concentrations of hydrocortisone 1% &
2.5%. How many grams will you use from 2
concentration

A. 20 gm from 1% and 40gm from 2.5%

B. 40gm from 1% and 20gm from 2.5%
C. 30gm from both
D. 10gm from 1% and 50gm from 2.5%
Answer:
Try the choices ration in the equation
(C1*V1)+(C2*V2)=(C*V)
(1 * 20) + (2.5 * 40) = (2* 60)
20 + 100 = 120
 Another Answer:
1% 2.5%
2%

0.5 + 1 = 1.5
X + Y = 60 g
0.5* 60/ 1.5= 20
1*60/ 1.5= 40
7- Prescription Hydrocortisone 2% w/w
Cold cream hydrocortisone solute 100mg/ml
how many ml will you use from this solution?

A. 10ml
B. 20ml
C. 40ml
D. 30 ml
Answer:
2% w/w= 2% 100gm
2gm means the prep. Needs 2 gm of hydrocortisone
100mg in 1 ml
0.1gm in 1 ml
2 gm in X ml
X= 1x 2/0.1= 20 ml
8- If we have 0.8687g CaCl2 in 500 ml solvent, density of the
solvent is 0.95 g/cm3. Find the Molality.
A. 0.0165 Molal
B. 0.0156 Molal
C. 0.0165m
D. 0.0156m
Answer:
Moles= Mass/ M.wt= 0.8687/ 111= 0.00782
Weight= Density x Volume= 0.95 x 500= 475gm= 0.475kg
Molality= moles/kg of solvent= 0.00782/0.475= 0.0165 Molal
9- How gm of substances X must added to 2000 gm
of 10% substances X solution in order to prepare
25% of substance x solution.

A. 10000 gm
B. 400 gm
C. 40 gm
D. 10 gm
E. 0.4 gm
Answer:
(C1 x V1) + (C2 x V2)= (C x V)

(100 x X-gm) + (10 x 2000 gm)= (25 x 2000+ X-gm)

100X + 20000= 50000 + 25X

100X – 25X= 50000 – 20000

75X= 30000
X= 30000/75
X= 400gm
Another answer:
100 10
25
15 + 75 = 90
X + 2000 = 2000+X

15*2000/75= 400
X= 400gm
10- How much water should be added to 250 ml of
1:500 w/v solution of benzalkonium
chloride to make a 1:2000 w/v solution?

A. 0.4L
B. 2L
C. 0.75L
D. 1L
Answer:
1:500 means 1g in 500ml
Xg in 250ml
X= 1*250/500= 0.5g in 250 ml

1: 2000 means 1g in 2000 ml

0.5g in Xml
X= 0.5*2000/1= 1000ml, 0.5g in 1000ml
So the water required is 1000-250= 750ml= 0.75L
11- How many Mosm are present in liter of Sodium
Chloride injection (M.wt: sodium chloride= 58.5)

A. 308 Mosm
B. 1026 Mosm
C. 456 Mosm
D. 234 Mosm
Answer:
Normally conc. of NaCl injection= 0.9%
That means 0.9 gm in 100 ml
That means 9 gm in 1 L

Mill moles= Wt (gm) / M.wt (gm) x 1000= 9/58.5 x 1000= 154

Milli moles= Wt (mg) / M.wt (gm)
Mosm = Milli moles x No. of dissosiation particles= 154 x 2= 308 Mosm
12- A solution contains 448 mg of KCl (MW=74.5)
and 468 mg of NaCl (MW= 58.5) in 500 ml. What
is the osmolar conc. of this solution?

A. 0.056 Osm/L
B. 0.56 Osm/L
C. 5.6 Osm/L
D. 56 Osm/L
Answer:
For KCI
0.448 gm in 500 ml
X gm in 1000 ml
Moles= 0.896/ 74.5= 0.012
Osm= Moles x no. of dissosiation particles= 0.012 x 2= 0.024
For NaCl
0.468gm in 500ml
X gm in 1000 ml X= 0.936 gm
Moles= 0.936/ 58.5= 0.016
Osm= 0.016 x 2= 0.032
Total Osmolar conc. of sol. = 0.032+0.024= 0.056 Osm/L
13- A patient weighting 80 kg is supposed to
receive a drug at a dose 2mg/ kg/ day. What
is the dose that the patient should take for
each day:
A. 80mg
B. 160 mg
C. 240mg
D. 320mg

2* 80= 160 mg
14- Drug X is a given to a 70 kg patient at an
infusion rate of 0.95 mg/kg/hr how much drug we
need for a 12 hr infusion bottle.

A. 798mg
B. 66.5mg
C. 665mg
D. 84mg

Answer: 0.95*70*12= 798mg

15- How many ml of water add to 5% KCL
solution to make 180 ml of solution (v/v)?

A. 171 ml
B. 234 ml
C. 230 ml
D. 9 ml
Answer:
5 ml in 100ml
X ml in 180ml
X= 5x180/100= 9ml
So, the amount of water is :180-9= 171 ml
16- Hypo parathyroid patient with tingling and numbness has
the following lab result so what is value of calcium
correlative to albumin when below 49.

Result Normal value

Calcium 1.6 2.25-2.6
Albumin 34 18-56
A. 2.3
B. 1.5
C. 2.5
N.B: 2.3 is constant value you have to know.
 Answer:
Ca - Alb
1.6 - 34
X - 49

1.6*49/34= 2.3
17- In a clinic patient prescribed with a 500 mg dose of
Aspirin, initial plasma conc. is 100 mg with half life 6
hours calculate total body clearance?

A. 0.5 L/hr
B. 5 L/hr
C. 50 L/hr
D. 500 L/hr
Answer:
Vd = Dose/ Initial concentration
= 500 / 100 = 5L
t1/2= 6hr

Cl= 0.693 Vd/ t1/2

= 0.693 x 5/6
= 0.5772 L/hr
18- Aminophylline (80%theophyline) was prescribed for asthmatic
patient in a dose of 500mg, half life= 6.93 hours how many hours will
it take to reach below 2%?

A. 42hr
B. 24hr
C. 18hr
D. 28hr

Answer:
(80%)…T1...(40%)…T2…(20%)…T3…(10%)…T4…(5%)…T5…(2.5%)…T6…(1.25%)

Time= 6 x t1/2= 6 x 6.93 = 41.5 hr

19- Drug Aminophylline (80% theophylline) in 500 ml
soln. Half life 6h. What is the conc. of Theophylline
after 1 day?
A. 5%
B. 10%
C. 15%
D. 20%
Answer:
1 day= 24 hr = 4 t1/2
(80%).….T1….(40%)……T2…..(20%)…..T3…..(10%)……….T4…….(5%)
20- For 1 liter of NaCl 3% calculate the Osmolarity
(M.wt= 58.5)

A. 1.026 Osm
B. 308 Osm
C. 1026 Osm
D. 10.26 Osm
Answer:
3% means 3gm in 100 ml
That means 30 gm in 1L
No. of moles= wt./ M.wt
=30/ 58.5 = 0.153 moles
Osm= No. of moles x No. of dissosiation particles
= 0.513 x 2= 1.026 Osm
Milliosmolarity = 1.026 x1000= 1026 Mosm
21- If we give 250 ml of a drug and the area under curve was
112mg/hr/ml. and after that we give 500 ml and the area
under curve was 56 mg/hr/ml. That bioavailability decreased
by

A. 25%
B. 50%
C. 75%
D. 100%
Answer:
250 ml………..112
500ml………….X
X= 112x500/250 = 224
But real AUC was=56
So the Bioavailability decreasing= 56/224x100 = 25%
22- Drug A taken IV and drug B taken orally the AUC of A= 300
and AUC of B=225. What is bioavailability of drug
A. 85%
B. 90%
C. 75%
D. 80%

Answer:
Bioavailability = AUC oral/AUC iv x 100
=225/300 x 100
= 75%
23- t1/2 in first order kinetics is

A. 1/k
B. 0.693/ k
C. Log K
24- A drug is given as iv infusion in a rate of
2mg/hr. its t1/2 = 2hr. How much mg of the
drug we need to reach steady state.

A. 4mg
B. 16mg
C. 20mg
D. 40mg
Answer:
We reach steady state after 5 t1/2
=5 x 2=10hr
2mg …for …1hr
Xmg …for…10hr
X= 2 x 10/1= 20mg
25- A drug with t1/2=72hr. the body will
receive complete dose after:
A. 1day
B. 2days
C. 1week
D. 2weeks
Answer:
We will reach steady state after 5 half-life
=5 x 72= 360hr = 15 days = 2 weeks
26- A patient takes levofloxacin 250mg/ml the pharmacist
has levofloxacin injection 500mg/ 20ml the concentration
needs to be dilated for patient. Which of the following
concentration is more accurate:
A. 10ml
B. 15ml
C. 7.5ml
D. 2 ml
Answer:
500 mg in 20ml
250mg in Xml
X=20 x 250/500=10ml
27- Prescription for a child contains Omeprazole syr. 10mg/ml
twice daily for a week, you have Omeprazole capsule 20 mg in
your pharmacy. How many capsules are needed to prepare
solution with concentration 2 mg/ml?

A. 7 cap
B. 8 cap
C. 9 cap
D. 10 cap
Answer:
10 mg/ml twice daily for a week=10*2* 7= 140
20 --------1 cap
140 ----- X cap
X= 1* 140/20=7 cap
28. Drug 500mg and 300mg eliminated outside the body and
t1/2=5hr and another drug same first one but with cone
1000mg. How many hours it take to eliminate 600mg out of the
body?

A. 5 hrs
B. 6 hrs
C. 8 hrs
D. 2 hrs
Answer:
Cls=Rate of elimination / drug conc

Cls1 = 300/500 = 0.6

Cls = 0.693* Vd / t1/2
Vd = t1/2 x Cls/0.693
=5x0.6/0.693
=4.3

Cls2=600/1000=0.6
t1 /2=0.693 x Vd/Cls
=0.693 x 4.3/0.6
=5 hrs
29. How can you prepare 100 ml of 12% MgCI by
taking

A-12ml of MgCl dissolve in 100 ml water

B-12 gm of MgCl dissolve in 100 ml water
C-12ml of MgCl dissolves in 1000 ml water
D-90.5 ml of MgCl dissolves in 100 ml water

Note; w/v = g/ml ex; 4% w/v means 4 gm in 100 ml

30- Man 40 years and 80 kg S.Cr - 0.5 mg/dl find
Creatinine clearance in mg/ml

A. 222 mg/ml
B. 232 mg/ml
C. 56 mg/ml
D. 75 mg/ml
Answer:
Cr. Cl for Male= (140-age)x weight/ 72 x ser. Creatinine
= (140-40) x 80/72 x 0.5= 55.55

NB: The same data for female the answer: 47

Cr.Cl for female = Cr.Cl for Male x 0.85
= 56 x 0.85= 47.22