Doc Laarni Calcu Flashcards by Lovely jane Luchavez

1- amount of drug is 5 mg in 1 ml what the amount of drug in 1
tsp in microgram
a ) 5
b ) 25
c ) 500
d ) 2500
e ) 25000

e ) 25000

Answer:
1 tsp = 5 ml
5 mg …. 1 ml
X mg …. 5 ml
X = 5 x 5/1 = 25 mg = (25 x 1000) 25000 mcg

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2- A solution is made by dissolving 17.52 g of NaCl exactly 2000 ml.
What is the molarity of this solution?
a- 3.33
b- 0.15
c- 1.60
d-3.00 x 10 -4
e-1.6x10 -4

b- 0.15

Answer :
Molarity=mole/volume (L)
1 Mole=molecular weight of subs. In 1 grams
No of Moles = wt / Mwt
So, molecular weight of NACL=23+34=57
So, Mole=17.52/57=0.307
So, Morality=0.307/2=0.153

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5ml of injection that conc. 0.4% calculate the amount of drug?
a-0.2mg
b-2mg
c-200mg
d-2000mg
e-20mg

e-20mg
Answer:
0.4 gm … 100 ml
X gm … 5 ml
X = 5 x 0.4/100 = 0.02 gm = (0.02 x 1000) = 20 mg

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An elixir contains 0.1 mg of drug X per ml. HOW many micrograms are
there in one tsp of the elixir
A. 0.0005 micrograms
B. 0.5 micrograms
C. 500 micrograms
D. 5 micrograms
E. 1500 micrograms

C. 500 micrograms

Answer :
0.1 mg in 1 ml
X mg in 5 ml
X = 0.1×5 /1 = 0.5 mg = 500 micro

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sol contain D5W another one contain D50W we want to prepare sol cotain
D15W its volune is 450ml … how much ml we need of each sol

a) D50w/D5w=10/35

Answer:
try the choices ratio in the equation :
(C1 × V1) + (C2 × V2) = (C × V)
( 50 × 10 ) + ( 5 × 35 ) = ( 15 × 45 )
Another answer :
(X) 50 ———- 10 15 – 5 = 10
15
(Y) 5 ———– 35 50 – 15 = 35
X / Y = 10 / 35 ———- Y = 3.5 X
X + Y = 450 ———- X + 3.5 X = 450
4.5 X = 450 ——— X = 450 /4.5 = 100
Y = 3.5 X = 3.5 x 100 = 350
X = amount of D50w …. Y = amount of D5w

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prescription
hydrocortisone 2%
Cold cream 60gm
You have concentrations of hydrocortisone 2.5% & 1% how many grams will
you use from two concentration?
a- 20gm from 1% and 40gm from 2.5%
b- 40gm from 1% and 20gm from 2.5%
c- 30gm from both

a- 20gm from 1% and 40gm from 2.5%

Answer:
try the choices ratio in the equation
( C1 × V1 ) + ( C2 × V2 ) = ( C × V )
( 1 × 20 ) + ( 2.5 × 40 ) = ( 2 × 60 )
Another answer :
(X) 2.5% ———- 1 2 – 1 = 1
2%
(Y) 1% ———– 0.5 2.5 – 2 = 0.5
X / Y = 1 / 0.5 ————– X = 0.5 Y
X + Y = 60 ——– 0.5 Y + Y = 60
1.5 Y = 60 ———– Y = 60 /1.5 = 40
X = 0.5 Y = 0.5 x 40 = 20
X = amount of 2.5 % …. Y = amount of 1%

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Prescription
hydrocortisone 2% w/w
Cold cream 60gm
you have hydrocortisone solu. 100 mg/ml .. how many milliliters will you use
from the solution ?
a.10 ml
b.20 ml
c.40 ml

b. 20ml

Answer :
2% w/w = 2% x 100gm = 2 gm means the prep. needs 2 gm of
hydrocortisone
0.1 gm in 1 ml
2 gm in X ml
X = 1 x 2/0.1 = 20 ml

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if we have 0.8687g cacl2 in 500 ml solvent , denisty of the solvent is 0.95
g\cm3 ….Find the molality
a- 0.0165 Molal
b- 0.0156 Molal
c- 0.0165 m
d- 0.0156 m

a- 0.0165 Molal

Answer :
Moles = mass/m.wt = 0.8687 / 111 = 0.00782
Weight = density × volume = 0.95 × 500 = 475 gm = 0.475 kg
Molality = moles / kg of solvent = 0.00782/0.475 = 0.0165 molal

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How gm of substance X must added to 2000 gm of 10% substance 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

b) 400 gm

Answer:
(C1 × V1) + (C2 × V2) = ( C × V )
( 100% × Xgm ) + ( 10% × 2000 gm ) = ( 25% × 2000+X gm )
100X + 20,000 = 50,000 + 25X
100X - 25X = 50,000 - 20,000
75X = 30,000 ….. X = 30,000/75 = 400 gm
Another answer :
100% ————– 15 25 - 10 = 15
25%
10% —————- 75 100 - 25 = 75
so the ratio between 100% : 10 % to reach 25% = 15 : 75
2000 gm —— 75
X gm ———— 15
X = 2000 x 15 / 75 = 400 gm

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10- How much water (in milliliters) 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/0.2L
D/ 0.05L

A/0.4L

Answer:
250/500 = 0.5
250/2000 = 0.125
0.5 – 0.125 = 0.375

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How many mOsm are present in 1 liter of sodium chloride injection
(Mwt: sodium chloride= 58.5) ?

308 mosm

Answer :
 Note ; normally conc. of NaCl injection = 0.9%
that means 0.9 gm in 100 ml ….. that means 9 gm in 1 L
 Step 1.
millimoles = wt (gm) / Mwt (gm) × 1000 = 9 /58.5 ×1000 = 154
Note ; millimole = wt (mg) / Mwt (gm)
 Step 2.
mOsm = millimoles x no. of dissosation particles =154 × 2 =308 mosm

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A solution contains 448 mg of KCl (MW=74.5) and 468 mg of NaCl (MW =
58.5) in 500mL. What is the osmolar conc. of this solution ?

0.056 Osm/l

Answer :
 For ( KCl )
0.448 gm in 500ml
X gm in 1000 ml …… X= 0.896 gm
moles= 0.896/74.5 = 0.012
Osm= moles × no. of dissosation particles =0.012 × 2= 0.024
 For NaCl
0.468 gm in 500 ml
X gm in 1000 ml ….. X= 0.936 gm
moles= 0.936 /58.5 = 0.016
Osm= 0.016 × 2= 0.032
 Total osmalar conc. of sol. = 0.032 + 0.024 = 0.056 Osm/l

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A Patient weighting 80 Kg is supposed to receive a drug at a dose of
2mg/kg/day. What is the dose that the patient should take for each day:
A. 80 mg ….. B. 160 mg ….. C. 240 mg ….. D. 320 mg ….. E. 400 mg

b. 160mg

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

A. 798 mg

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how many gm of water add to 5% KCL soln to make 180 gm of
solution(w\w)?

171 gm
Answer:
5gm————–100
Xgm————–180
X= 5x180/100=9 gm
So, the amount of water is:- 180 - 9 =171 gm

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hypoparathyroid patient with tingling and numbness has the following lab
result so what is value of calcium correlative to albumin when below 45

Calcium
Result: 1.6
normal value: 2.25-2.6

Albumin
Result: 34
normal value : 18-56

a. 2.3
b. 1.5
c. 2.5

a. 2.3

N.B: 2.3 is a Conistant value you have to know

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in clinic patient prescriped with a 500mg dose of aspirin , initial plasma
conc is 100mg .. With half life 6 hours calculate total body clearance ?
a.0.5 L/hr
b.5 L/hr
c.50 L/hr

a. 0.5 L/ hr

Answer:
Vd = dose / initial conc = 500/ 100 = 5L ….. T1-2 = 6 hr
Cl = 0.693 Vd / T1-2 = 0.693 × 5 / 6 = 0.5775 L/hr

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aminophylline (80%theophylline) was prescriped for asthmatic patient in
a dose of 500mg , half life =6.93 hours how many hours will it take to reach
below 2 % ?

42 hr
Answer:
(80%) …T1… (40%) …T2… (20%) …T3… (10%) …T4… (5%) …T5… (2.5%)
…T6… (1.25%)
Time = 6 × T1/2 = 6 × 6.93 = 41.5 hr

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Drug aminophylline (80% theophylline) in 500ml sln . Half life 6 h .what is
the concn of theophylline after 1 day ?

5%
Answer:
1 day = 24 hr = 4 T1-2
(80%) ….T1… (40%) …T2… (20%) …T3… (10%) …T4… (5%)

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For 1 litre of NaCl 3% calculate the osmolarity m.wt=58.5

1026
Answer:
3% means 3gm in 100 ml … that means 30gm in 1L
No. of moles = wt / Mwt = 30 / 58.5 = 0.513 mole
Osm = no. of mole × no. of dissosation particles = 0.513 × 2 = 1.026
1.026 x 1000 = 1026 mosm

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If we give 250 ml of a drug and the area under curve was 112mg/hr/L and
after that we give 500 ml and the area under curve was 56 mg/hr/ml
The bioavilability decreased by
A-25%
b-50%
c-75%

A-25%
Answer:
250ml …….. 112
500 ml ………X
X= 122×500 / 250 = 224
But real auc was = 56
So the bioavilability decreasing = 56/224 ×100 = 25%

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drug A taken IV and drug B taken orally
the AUC of A =300 and Auc of b =225
what is biovalbility of drug
A. 85%
B. 90%
C. 75%
D. 80%

C. 75%

Answer:
Bioavailability= auc oral /auc iv ×100 = 225/300 × 100 = 75%

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T 1/2 .. in frist line is ….
A .1/k
B . 0.693/ k

B . 0.693/ k

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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

C .20mg
Answer :
We reach steady state after 5 T1-2 = 5 × 2 = 10hr
2mg …ever… 1 hr
Xmg …after… 10 hr
X = 2×10/1 = 20mg

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a drug with T1/2 = 72hr , the body will recive complete dose after ; A. 1 day B. 2days C. 1week D. 2weeks

D. 2weeks Ans: We will reach Steady state after 5 half-life = 5×72= 360hr = 2weeks

A patient takes levofloxacin 250mg/ml , the pharmacist has levoflaxacin injection 500mg / 20 ml , the concentration needs to be dilated for patient .. which of the following concentration is more accurate: A/ 10 ml B/ 15 ml C/ 7.5 ml

a. 10ml Answer : 500 mg in 20 ml 250 mg in X X = 20 x 250 / 500 = 10 ml

priscription for a child contain Omeprazol syr. 10 mg/ml twice daily for a week .. you have Omeprazol capsul 20 mg in your pharmacy, how many capsules are needed to prepare solution with concantration 2 mg/ml ??

7 cap. Answer: 10 mg/ml twice daily for a week = 140 20 _____1 140 _____ X X=140/20=7

Drug 500mg and 300mg eleminated outside the body and t1/2=5hr and another drug same first one but with conc 1000mg .. how many hrs it take to eliminate 600mg ot of the body?

5 hrs Answer : CLs=rate of elimination /drug conc CLs1=300/500=0.6 Vd=t1/2×cls/0.693=5×0.6/0.693=4.3 CLs2=600/1000=0.6 t1/2=0.693×vd/cls=0.693×4.3/0.6=5 hrs

HOW can prepare 100 ml of 12% MgCl 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 dissolve in 1000 ml water d-90.5 ml of MGCL dissolve in 100 ml water

b-12 gm of MGCL dissolve in 100 ml water Note ; w/v = g/ml ..... ex ; 4% w/v means 4 gm in 100 ml

man 40 years and 80 kg sr ce 0.5 mg\dl find creatinie clearance mg\ml : a.222 b.232

a.222 Answer : Cr.cl for male = (140 – age)x weight /72 x ser. Creatinine =(140 – 40 ) x 80 / 72 x 0.5 = 222 N.B : The same data for female the answer is : 189 Cr.cl for female = Cr.cl for male x 0.85 = 222 x 0.85 = 188.7

31.15 g of drug is added in 150mg of a solvent. Then what is the total concentration of drug in the final mixture: a- 6.01% b- 9.10% c- 10% d- 15%

b- 9.10% Answer: 15 + 150 = 165 15 g in 165 X g in 100 X = 100 x 15 / 165 = 9.10

A bag containing 250 ml of 25000 IU heparin The patient weigh 70 kg should recieve 10 IU/kg/hr...calculate the amount in ml the the patient should recieve in one hour..

7 ml Answer: 10 iu for 1 kg X iu for 70 kg X = 70 x 10 /1 = 700 iu 250 ml of 25000 iu X ml of 700 iu X = 700 x 250 / 25000 = 7 ml

Patient with prescription of Captopril 50 mg per tab with a dose of 100 mg daily for 4days and you only have the 25 mg tab .. How many tablets you will dispense ?

16 Tab Answer : 100 mg daily for 4 days = 400 mg 400/25 = 16 tab

A problem with the following data Dose = 1000 Initial conc =10 Elimination rate constant=0.1 Calculate total clearance ?? a-250 b-200 c-150 d-100 e-10 litre

e-10 litre Answer: Cl= vd × kel Vd=dose/conc=1000/10=100 Cl=0.1×100=10

Problem with the following data : Density = 1.75 g/cm3 Mass = 15 gm Calculate the Volume ? a.11 b.10 c.8.52

c. 8.52 Answer: Denisty = mass / volume volume = 15 / 1.75 = 8.57

Prescription contain : Clindamycin 1.5% dilultion with alcohol up to 300 ml you have a bottle 100 ml of 10% clindamycin how many millelitres will you use ? a.7.5 b.45

b.45 Answer: 1.5 ..... 100 X ..... 300 X=4.5 10 ...... 100 4.5 ..... X X=45

A drug with Conc. 400 m and T1/2 = 12 hr.s the concentration will decrease after 1 day by … a.10% b.25% c.75% d.90%

c. 75 % Answer: 24 hr.s = 2 half lives (400) …T1 … (200) … T2 … (100) so you lose 300 of the drug ( 300 / 400 ) x 100 = 75%

A drug should be given 50 ml of 2 meq/ml , but available concentration is 10 meq/ml, How many ml should dispense to patient? a.5 ml b.10 ml c.15 ml d.20 ml e.25 ml

b.10 ml Answer: 2mg -----1ml X mg-----50ml X = 50 x 2 =100ml 10 mg-------1ml 100 mg------ X X = 100 x 1 / 10 =10 ml

30gm of 1% hydrocortisone mixed with 40 gm 2.5% hydrocortisonen what is the concentration of the resulting solution? a) 3% b) 1.85% c)10% d) none of the above

b) 1.85% Answer : C1.V1 + C2.V2 = C3.V3 30gm × 1% = 0.3gm 40gm × 2.5% = 1gm So, 1.3 gm is in 70 gm So, the con. =1.3/70=1.857%

if we have 90% of substance X solution , 50% of substance X solution , how mixing both to give 80% of substance X solution ? a- 3 : 1 b-1:3 c-10:30 d- 5:9

a. 3:1 Answer : We should try all answer with that equation (C1×V1) + (C2×V2) = (C×V) (90% × 3) + (50% × 1) = (80% × 4) ( 270 ) + ( 50 ) = ( 320 ) ( 320 ) = ( 320 ) so the answer is 80% Another answer : 90% 50% 80% 30 10 So .. 90/50 to reach 80 % equal 30/10 = 3/1

prep. contain coal tar 30 part ... petroleum 15 part ... adeq. to 150 part ... what conc. of coal tar in 500 ml:

100 part Answer: 30 part present in 150ml of prep. X part present in 500ml of prep so, conc. of coal tar in 500ml=30x500/150= 100 part

42.How many grams needed from drug in one teaspoonful , if 5 tspfull doses contain 7.5 gm of drug ? a) 0.0005 b) 0.5 c) 500 d) 1.5

d) 1.5 Answer: 7.5gm in 5 tsp ..... X gm in 1 tsp ..... X = 7.5×1 /5 = 1.5 gm N.B: 1 tsp = 5 ml

KI solu. has 0.5mg/ml dissolve in 30ml water calculate the amount of KI in the solu. ?

15mg Answer : 0.5 mg in 1 ml X mg in 30 ml X= 0.5×30 /1 = 15 mg

the dose of drug is 0.5ml per day and the total amount of the drug Is 100ml what is the total dose ?

200 Answer : no. of doses = amount of drug / amount of one dose = 100/0.5= 200

if we have a solvent costs 150 riyal/kg and its specific gravity =1.07 ,so the cost for 100ml of the solvent is :

16.05 riyal Answer : Weight (Kg) = volume (L) × sp. Gravity …….. 100 ml = 0.1 L wt = 0.1 × 1.07 = 0.107 Kg 1 kg cost 150 riyal 0.107 kg cost X riyal X = 0.107×150 /1 = 16.05 riyal

A patient cholesterol level is equal to 4mM/L. This cholesterol level can be expressed in terms of mg/dL ( molecular weight of cholesterol = 386) A.0.0154 mg/dL B. 0.154 mg/dL C. 1.54 mg/dL D. 15.4 mg/dL E. 154 mg/dL

E. 154 mg/dL Answer : Conversion from (mM) to (mg) = conc. × molecular weight Conversion from (L) to (dL) = conc. / 10 Conc (mg/dl) = conc. (mMol /L) × mwt / 10 = 4×386 /10=154.4

drug container contain 90 mg each tablet contain 0.75mg. how many doses ?

? No. of doses = total wt / wt of one dose = 90 / 0.75 = 120 dose

How need prepare benzacainamid conc. 1:1000 ,30cc of benzocainamid solution? a-30 mg b-50 mg c-80 mg d-100 mg e-130 mg

a. 30 mg Note : cc = cubic centimeter = cm3 = ml Answer : 1 gm ----- 1000 ml X gm ----- 30 ml X = 30 x 1 / 1000 = 0.03 gm = 30 mg

The Molal concentration of 0.559 M solution is ; (Mwt=331.23 g/mol) (density of solution =1.157g/ml) a-1.882 b-0.882 c-0.559 d-0.575

d-0.575 Answer : Mass = moles × Mwt = 0.559 × 331.23 = 185.15 gm wt of solution = Volume × Destiny = 1000 ml × 1.157=1157 gm so wt of solvent = 1157 - 185.15 = 971.85gm = 0.971 kg molality = moles / kg of solvent = 0.559 / 0.971= 0.575 molal

Problem asked to calculate Plasma Osmolarity an you have given some data Na 140 Cl 103 Hco3 18 Bun 8 S.cl 8

Answer is : 263 N.B:  the data of this problem isn't complete here .. 263 is the right answer just know it  in general .. to calculate plasma osmolarity follow this equation : 2[Na] +[Glucose]/18 +[BUN]/2.8

ug decrease after 2hr to 50% &the user takes it every 2 hr how many hours needed to reach steady state ? A/2-4 B/6-8 C/10-12

C/10-12 Answer: Time to reach steady state ((Tss)) = 4 to 5 T1/2 4 x 2 = 8 ….. 5 x 2 = 10 N.B: if there is (( 8-10 )) if choices … choose it

10g of a drug was dissolved in 150g of solvent, what is the final concentration?

6.25% Answer: 10 … 160 X …. 100 X = 100 x 10 / 160 = 6.25 %

A physician prescribed paracetamol 120mg/5ml to take 10ml every 8 hours but the pharmacist has only paracetamol 160mg/5ml . what is the volume to be administered to give the effect of the first dose : a- 6.5 ml b- 7.5 ml c- 10 ml d - 11 ml

b- 7.5 ml Answer: dose = 240 mg paracetamol 160 mg in 5 ml 240 mg in X ml X = 240 x 5 / 160 = 7.5 ml

A drug with conc. 100 mg/ml .. after 1 hr. it decreased to 50 mg/ml .. calculate its concantraion after 3 hours : a.25 b.12.5 c.6.25

b.12.5 Answer : 100 .. [1hr] .. 50 .. [2hr] .. 25 .. [3hr] .. 12.5

how many gm of water add to 5% KCL soln to make 100 gm of solution (w\w) ?

95gm N.B: 5% (w/w) means 5gm of KCl in 95gm of water and solution total wt=100

1000 mg of drug follow one compartment.. calculate vd ? Time conc 0 hr 80 2 hrs 58 4 hrs 34 6 hrs 28 12 hrs 10 A.12.5 litre B. 4 litre C. 45 litre

A.12.5 litre Answer : Vd = dose / initial conc. Vd = 1000 / 80 = 12.5 L

57. Drug dose 1000 mg orally Time concentration 0 hr 40 2 hr.s 18 4 hr.s 8 What is the Vd of the drug ? a.55 litre b.45 litre c.75 litre c

d.25 litre Answer: Vd= 1000/40 = 25 L

HOW can prepare 100 ml of 12% MgCl 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 dissolve in 1000 ml water d-90.5 ml of MgCl dissolve in 100 ml water e-0.95 ml of MgCl dissolve in 100 ml water

b-12 gm of MgCl dissolve in 100 ml water

59. How many grams of drug used to prepare 150 ml solution ,, if one tsp contains 7.5 mg of drug a. 4 gm b. 0.225 gm c. 2.25 gm

b. 0.225 gm Answer: 7.5 mg in 5 ml X mg in 150 ml X = 150 x 7.50 / 5 = 225 mg = ((225/1000)) 0.225 gm

Patient takes dose 20 mg/kg/day what is the dose if patient weight 60 pound ?

545 mg/day Answer: you have to know .. 1 kg = 2.2 pound (lb) 20 mg -------- 2.2 lb X mg ---------- 60 X = 60 x 20 / 2.2 = 545.45 mg/day

61.A child was prisciped a drug with dose 65 mg/kg/hr .. his body weight = 35.2 pound Calculate the dose .. a.1.040 gm b.10.40 gm

a.1.040 gm Answer: 35.2 pound = 15.97 kg = about 16 kg 65 mg … 1 kg X mg … 16 kg X = 16 x 65 = 1040 mg = 1.040 gm

Calculate the Specific gravity of a substance of volume = 121.92 ml & wt = 107.5 A/1.88 s.g. B/2.88 s.g. C/0.88 s.g. D/8.8 s.g.

C/0.88 s.g. Answer: Denisty = wt. / volume = 107.5 / 0.12192 = 881.7 Sp. Gravity = denisty Of substance / den. Of water = 881.7 / 1000 = 0.88

The ppm concentration of a 6.35x1 0-6M solution of sucrose (Mwt of sucrose is 342.3 g/mole) is: A. 2.174 × 10-3ppm B.2.174 ppm C.2.174 × 10-6 ppm

B.2.174 ppm ppm concentration = mass in mg / volume in liters Molar conc means no. of mole in 1 liter .... then volume= 1L mass = moles × Mwt = 6.35x10-6 x 342.3 = 2.174x10-3gm = 2.174 mg Then 2.174 mg is in 1L = 2.174 ppm

A 500 infusion bottle contains 11.729 mg of potassium chloride (KCI). How many mEq of KCI are present? ( Mwt of KCI = 74.6) A. 0.1571 mEq B. 1571 mEq C. 6.37 mEq D. 0.00637 mEq

A. 0.1571 mEq Answer : mEq = wt (mg) × valency / Mwt = 11.729 ×1 / 74.6 mEq = 0.1572

Fifty micrograms equals: a-50000 ( nanogrames ) b- 0.05 ( milligrams ) c- 0.0005 g d- a and b e- a and c

d- a and b Note; ... mc-g = 1000 nano-g ... milli-g = 1000 mc-g ... g = 1000 mg

a 2 mg/L solution , according ppm a-2 ppm b-0.002 ppm c-0.000002 ppm

a-2 ppm Note ; ppm = mg / L ppm : part per milion

What is The Specific gravity of substance has Weight=Y & The volume is X ?

Y/X Answer : The Specific gravity =Density of the substance/Density of water Density of water = 1 .... Density of substance = weight/volume So, the sp. gravity of sub. =weight (Y) /volume(X)/1 = Y/X

drug decrease to 50% of its plasma conc. after 2hr .. we have dose A given each 2hr and dose B given each 4 hour … in dose B what is the plasma conc. at steady state ? A/0.25 B/0.5 C/2

B/0.5

Calculate C av .ss 1gm vancomycin for patient 78 kg Taken by infusion rate 12 hr /7 day T 1/2 =8 Vd = 1 k/l A. 3 B. 5 C. 17 D.19

We can't find the right answer .. try to solve it 

Paitents on treatment with acyclovir and famcyclovir .. group that treated by acyclovir show recurrence by 27% and who treated by famcyclovir show recurrence by 25% the ques. is how many patients should take famcyclovir over than who take acyclovir per year to reach equivilant results ?

The answer is : cannot be calculated because of low information

Patient's dose of some drug is 0.5 mg daily and Vd = 500 L .. his body elimination rate is 110.16 Litre per day … in the last day about 80 % of the drug was in his blood Calculate half life ..

3days Answer: Cl=0.693 x vd / T1.5 T1/2 = 0.639 x 500 / 110.16 = 3.14 day

Problem with data : drug 10 mg/ml and t1/2= 3 hrs how much hrs needed to reach steady state??

12 – 15 Answer: Time required to reach steady state (Tss) = 4 – 5 t1/2 4x3=12 ……. 5x3=15

drug t1/2= 2h .. dose A taken every 2h and dose B taken every 4h compare plasma concentration a to b .. a.1/2 b.2

b.2

A half life of a drug decrease by 50% , after how hours will the time needed to decrease to 2% a.2 …. b.10 …. c.5 …. d.12

d. 12 Answer : 100% .. [T1] .. 50% .. [T2] .. 25% .. [T3] .. 12.5% .. [T4] .. 6.25% .. [T5] .. 3.1% .. [T6] 1.5% so we need 6 half lives to reach below 2% ......... T1/2 = 2 h. 2 x 6 = 12 h.

A problem with thin curve and ask for therapeutic range

answer : 8/2 = 4 - in other exams the same curve with LD50 = 20 & ED50 = 5 so TI = LD50/ED50 = 20/5 = 4 page 28

which drug has higher bioavailability ? 1.A … 2.B … 3.C … 4.D page 29 drawing

1.A N.B : bioavilability measured by comparing plasma level higher plasma level = higher bioavailability

Which drug of the following has the safest margine ? 1.A … 2.B … 3.C … 4.D picture at page 30

1.A

Molarity of 17.52 NaCl solution

0.15

Cold cream with two concentrations :

20gm from 1% and 40gm from 2.5%

.Cold cream (( how many ml uses ))

20mL

Ca correvted to albumin :

2.3

Osmolarity of NaCl

1026

AUC bioavailability ((112, 500))

25%

AUC bioavailability ((300, 225))

75%

Levofloxacin

10mL

omeprazole

7 cap

Crcl of Male, 40 y, 80 kg with Scr: 0.5 mg/dL

222mL/min

Crcl of Male, 40 y, 80 kg with Scr: 0.5 mg/dL the same problem but for female :

189mL/min

heparin bag

7mL

captopril

16 tablets

clindamycin

plasma osmolarity

263

paracetamol

7.5mL

gm of water add to 5% KCL (( w/w )) :

95 gm