5 Flashcards
memorize the formulas
formula of Molality
moles / kg of solvent
formula of molarity
moles / L of solvent
formula of /MMoles
wt g
_____
M.W
formula or Meq
wt mg x verelance
____________
M.W
formula of osmolarity
wt g/L
________
M.W
X practice
look at page 2
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
PJ«PO
;PJ«PO
X = 5 x 5/1 = 25 mg = (25 x 1000) 25000 mcg
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
3-5ml 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:
0.4gm …100ml
X gm…5mL
X = 5 x 0.4/100 = 0.02 gm = (0.02 x 1000) = 20 mg
4-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
5- 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
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6- 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
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7-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.20 ml
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
8- 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
- 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
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
11-How many mOsm are present in 1 liter of sodium chloride injection
(Mwt: sodium chloride= 58.5) ?
308 mosm
Answer :
x Note ; normally conc. of NaCl injection = 0.9%
that means 0.9 gm in 100 ml ….. that means 9 gm in 1 L
x Step 1.
millimoles = wt (gm) / Mwt (gm) × 1000 = 9 /58.5 ×1000 = 154
Note ; millimole = wt (mg) / Mwt (gm)
x Step 2.
mOsm = millimoles x no. of dissosation particles =154 × 2 =308 mosm
12-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 :
x 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
x 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
x Total osmalar conc. of sol. = 0.032 + 0.024 = 0.056 Osm/l
- 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. 160 mg …..
- 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 …..
- 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
- 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
- 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
- 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 % ?
- aminophylline (80%theophylline) was prescriped for asthmatic patient in
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
20.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
- 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%)
- 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%
- 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%
23.T 1/2 .. in frist line is ….
A .1/k
B . 0.693/ k
B . 0.693/ k
- 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
- 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/ 10 ml
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
28.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 mg 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
33.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
34.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
e-10 litre
Answer:
Cl= vd × kel
Vd=dose/conc=1000/10=100
Cl=0.1×100=10
35.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
36.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
37.A drug with Conc. 400 m and T1/2 = 12 hr.s
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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:
- prep. contain coal tar 30 part … petroleum 15 part … adeq. to 150 part …
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
43.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 ?
- the dose of drug is 0.5ml per day and the total amount of the drug Is
200
answer:
no. of doses = amount of drug / amount of one dose = 100/0.5= 200
45.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 ….100ml=01.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
46- 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
47.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
48- 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
- drug 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 «[
N.B: if there is (( 8-
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50.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:
x the data of this problem isn’t complete here .. 263 is the right answer
just know it
x in general .. to calculate plasma osmolarity follow this equation :
2[Na] +[Glucose]/18 +[BUN]/2.8
10g of a drug was dissolved in 150g of solvent of solvent, what is the final concentration ?
6.25%
answer:
10….160
x….100
x=100 x 10/160=6.25%
54.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
53.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
how many gm of water add to 5% KCL soln to make 100 gm of solution (w)(w)?
95 gm
N.B: 5% (w/w) means 5 gm of KCl in 95 gm of water and solution total wt=100
drug dose 1000mg orally
time 0hr
concentration :40
time:2hrs
concentration:18
time 4hrs
concentration:8
what is the Vd of the drug?
a. 55 liter
b. 45 liter
c.75 liter
d. 25 liter
d. 25 liter
answer:
Vd=1000/40=25L
1000 mg of drug follow one compartment. calculate vd?
time :0hr
concentration mg/ ml : 80
time :2hrs
concentration mg/ml:58
time:4hrs
concentration mg/ml:34
time:6hrs
concentration mg/ml:28
time :12hrs
concentration mg/ml:10
a. 12.5 liter
b. 4 liter
c. 45 liter
a. 12.5 liter
answer:
Vd=dose/ initial concentration
Vd= 1000/80=12.5L
- 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
patient takes dose 20 mg /kg/day
what is the dose if patient weigh 60 pounds ?
545 mg/ day
answer:
you have to know.. 1kg =2.2 pound (lb)
20mg—-2.2lb
xmg—60
x=60 x 20 /2.2= 545.45 mg/day
- 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
a child was prisciped a drug with dose 65mg/kg/hr… his body weight = 35.2 pound
calculate the dose ..
a. 1.040gm
b. 10.40 gm
a. 1.040 gm
answer:
35.2 pound =15.97kg= about 16 kg
65mg…1kg
xmg…16kg
x=16x65=1040mg=1.040gm
calculate the specific gravity of substance of volume =121.92ml and wt =107.5
a. 1.88s.g.
b. 2.88 s.g.
c. 0.88 s.g
d. 8.8 s.g
c. 0.88 s.g
answer:
density= wt./volume
=107.5/0.12192=881.7
sp. gravity= density 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
Answer :
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 (KCl). How many mEq of KCl are present? (Mwt of KCl = 74.63)
a.0.1574 mEq
b. 1571 mEq
c/ 6.37mEq
d. 0.00637 mEq
a.0.1574 mEq
answer:
mEq=wt(mg) x valenxy /Mwt=11.729x1/74.6 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:…mcg=1000 nano-g…milli-g=1000mc-g…g=1000mg
what is the specific gravity of substance has weight= y and the volume is x?
Y/X
answer:
the specific gravity= density of 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
a 2mg/L solution, according ppm
a. 2ppm
b. 0.002 ppm
c. 0.000002ppm
a. 2ppm
note:ppm=mg/L
ppm:part per million
69.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 -
patient’s dose of some drug is 0.5 mg daily and Vd =500 L..his body elimation rate is 110.16 liter per day … in the last day about 80% of the drug was in his blood
calculate half life…
3 days
answer:
Cl=0.693 x vd/T1.5
T1/2=0.639 x500/110.16=3.14 day
problem with data:
drug 10mg/mL and T1/2=3 hrs
how much hrs needed to reach steady state ??
12-15
answer:
time required to reach steady state (Tss)( =4-5t1/2
4x3=12………5x3=15
drug t1/2= 2h .. odse a taken wvery 2h and dose b taken every 4 h compate 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 hlaf lives to reach below 2% …………..T1/2 =2 hours .
2x6 =12h
which drug of the following has the safest margine?
1.A
2.B
3.C
4.D
1.A
N.B: safest margine = higher therapeutic index
look at page 31
Molarity of 17.52 NaCl solution :
0.15
75.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
look at page 29
76.which drug has higher bioavailability ?
1.A
2.B
3.C
4.D
1.A
N.B: bioavailability measured by comparing plasma level higher plasma level= higher bioavailability
page 30
cold cream with two concentreation
20 gm from 1% and 40 gm 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
10 mL
omeprazol
7 cap
Crcl of male, 40 y, 80 kg with Scr:0.5 mg/dL
222mL/min
the same problem but for female:
189 mL/min
heparin bag
7 mL
clindamycin
45
captopril
16 tablets
plasma osmolarity
263
paracetamol
7.5mL
gm of water add to 5% KCl ((w/w))
95 gm
what is the formula of osmolarity
wha is the formula of mosmol
what is the formula of CL
what is the formula of VD
what is the formula of T1/2
what is the formula of time to reach ss
what is the formula of PPM
what is formula of bioavailability
wha tis the formula of BSA
what is the formula of specific gravity
what is the formula of density
what is the formula of IBW
what is the formula of child BSA
what is the formula of CrCl
what is the formula of BMI
what is the formula of child dose
not yet done page 3-5
