can any one slove this calculation

[romi1011]

can anyone answer these questions? please

1. The half life of a drug is 9 days. a single 0.5 mg dose of the drug yields an [AUC] value of 408 ng h/ml. in nanograms per ml, what plasma level will resuly at steady state if this product is given once daily and is 77% bioavailable?

a. 5
b. 13
c. 17
d. 23

Ans: Css=AUC/T = 408/24 = 17
how will the value of T be 24?

2. A solution initially contains methyl acetate(0.01 M) and NaOH (0.01M). the solution is unbuffered and both reacting species are consumed. if the rate constatnt for this reaction at 25 degrees C is 1.082 litres/(mole.min), hoe many minutes will it take for the concentration of methly acetate to fall to 0.0090 M?

a. 92.4
b. 33.6
c. 10.3
d. 0.10


1/c-1/c0 = kt


Six hours after 500 mg of a drug is administered by IV injection, a patient's plasma concentration Is 10 microg/ ml .

If T 1/2 = 4 h
( MEC ) = 2 micorg/ ml

How many hours after the first dose should a second dose be administered ?


The radioactive decay of radium has a rate constant of 0.000422 per year. How many years will it take for 20% of the radium initially present to degrade? (The amount initially present can be specified as 100%.)

A. 230
B. 529
C. 1000
D. 3815
The answer is B (529).

[srilak]

Hi,


Six hours after 500 mg of a drug is administered by IV injection, a patient's plasma concentration Is 10 microg/ ml .



If T 1/2 = 4 h
( MEC ) = 2 micorg/ ml

How many hours after the first dose should a second dose be administered ?
Here,plasma con is 10micro/ml
1/2 life 4hrs

so 10 mcg/l----------after one half-------5mcg/ml--------time-4hrs
after secoond half----2.5mcg/ml-------another 4hr
then we cann't go for another half because the Mec conc is 2mcg/ml

So the next dose will be after 8hrs.


The radioactive decay of radium has a rate constant of 0.000422 per year. How many years will it take for 20% of the radium initially present to degrade? (The amount initially present can be specified as 100%.)

A. 230
B. 529
C. 1000
D. 3815
The answer is B (529).

T=2.303/k*loga/a-x

Here a-x wiil be =100-20=80%

T=2.303/0.00422*log100/80
T=528.8years

I will try another questions too.I looked these question answers in Mannonsheroff question book....but in that they answerd the questions very vaguly.

regards


[/quote]

[srilak]

Hi,

In first question, we are giving drug for once daily.

K we calculate in hours so T will be 24 hrs.

T=24hrs.

May be I am wrong.Let me know if u find this differently.

regards

[jruerph]

Yeah, that is a tough one

[srilak]

HI

2nd question I solved like this.

In mannonsheroff question book,they gave one fomula for this,even I don't know how to get that formula.

K=x/t*a*(a-x).

Here k= rate constant

X=consuming concentration=(.01-.009=.001)

a= intial concentration=.01
a-x=remaining concentration.=.009
k=1.082lit/min

T=.001/1.082*.01*.009
T=10.26min

Answer is C


Regards