Doc Laarni Calcu Flashcards
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
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
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
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
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
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%
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
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
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
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
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
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 % ?
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
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%)
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
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%
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. 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
- 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
- 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
45
plasma osmolarity
263
paracetamol
7.5mL
gm of water add to 5% KCL (( w/w )) :
95 gm
