calculations

[kvr]

can anyone solve these problems?

1. For children, streptomycin is to be administered at a dose of 30mg/kf of body weight daily in divided doses every 6 to 12 hours. The dry powder is dissolved by adding water for injection, in an amount to yield the desired concentration as indicated in the following table(for a 1-gram vial)

Approximate conc(mg/ml) ----volume(ml)

200----------------------------------- 4.2

250 ----------------------------------3.2

400 ---------------------------------1.8
reconstituting at the lowest possible concentration, what volume (ml) would be withdrawn to obtain one days dose for a 50 pound child?



2. how many milliosmoles of calcium chloride(cacl2.2H2O mol wt=147) are represented in 147 ml of a 10%w/v calcium chloride solution?



3. how many gms of talc should be added to 1 lb of a powder containing 20 gm of zinc undecylenate per 100 gm to reduce the concentration of zinc undecylenate to 3 %?

[fpgee_05]

Hi,

For ur Q2,
I got answer like this. Please confirm the answer.......

Osmole= mole (gram mol.wt)/i(dissociation factor)
i=1 for nonelectrolytes

But here CaCl2 2H2O hence i=2.6 (please refer shargel calculations chapter)
1 osmole = 147/2.6 = 56.5 g
1 m.osmol = 56.5 mg

10% w/v solution means 10 g/ 100 ml, hence 14.7g in 147 ml
Therefore we can figure out that # of milli osmoles as 260 as we know 1 m.osmol = 56.5 mg.

Please let me know whether this is the right answer if you have any other answer source..
Thnaks n Good luck to all of us..

For Q3:

I got from this forum only.. hope this helps..

This problem can be solved by Alligation method.
We know that talc is diluent (0%)
Zinc is the active ingredient, and the conc. of Zn in given powder is 20%w/w (i.e., 20gm Zn per 100 gm of powder)
We are given 1lb of the 20% mixture of Zn and talc. 1lb = 454gm, right?
So we wanna make the conc of Zn to 3% by adding more diluent, talc. And we are asked how much talc should be added for the same.
Using alligation method, put 0% on top, 20% below that and put 3% in middle, then find differences by alligation, u get on the right hand side,
17 parts of talc + 3 parts of 20% powder gives 20 parts of 3% powder.

i.e., 17gm of talc + 3 gm of 20% powder gives 20 gm of 3% powder.

Therefore, calculate the amount of talc to be added to 454gm of 20% powder to reduce the conc of zinc to 3%.

By calculation it can be found that
385.9gm talc + 68.1 gm of 20% powder gives 454gm of 3% powder mixture.

There's only one thing I'm not sure, does the question want us to calculate the talc to be added to the whole 20% powder (454gm), or just in what proportions we should mix talc with the given powder to get 3% powder mixture. If the question is for the latter, then I hope the above is the answer.
If the former is to be found, we should calculate the the amount of talc to be added to 454gm of 20% powder to reduce concentration into 3%.

385.9gm talc ? gm of talc
------------- = -----------------
68.1gm 20%pdr 454gm of 20% pdr


Therefore, ? gm of talc = (385.9 * 454)/ 68.1 = 2572.66gm

i.e; 2572.66gm talc should be added to 454gm 20% talc-Zn powder to get final powder of 3% Zinc conc.

Please tell how to solve Q1..

Thanks in advance

[jane2238]

FPGEE 05 you have confused dissociation constants which are needed in isotonic solutions

osmotic pressure is determined by the # of particles produced by the solution. For non electrolytes the # is 1, for electrolytes it is the # of particles produced in the solution and is influenced by the degree of dissociation. For ELECTROLYTES ASSUME YOU HAVE COMPLETE DISSOCATION.

So in this example the # of particles(species) in CaCL2 is 3

mOsmol/l = wt/mw x # of species x 1000

= 14.7/147 x 3 x 100


= 300mOsm

Be careful with these calculations because previous questions posted involve solutions that only partly dissociate so you have to figure out the mosmoles in the percentage which dissociates and use the # of species and the mosomoles in the percentage which doesn't dissociate which you treat as a non electrolyte so hence the # of species is 1. ( I'm not trying to confuse you but I've been involved with various groups for 2 years now and find the same questions being posted and I know what calculations tend to confuse people.)

I hope this helps, and that my answer is correct and that I didn't confuse you!

Jane

[fpgee_05]

Hi Jane,

Thanks for the correction..I was confused as I studied m.osmol concept from morris cody..
From shargel, yours is correct..\ I will follow shargel...
Tell me how to solve Q1 if you can

Thank you
Deepa

For Q1:

50 lb child = 22.72 kg
dose = 30mg/kg hence for this child 681mg/dose..
The lowest concentration we need hence I divided 681 mg /200 mg/ml, which gave 3.4

Just now I got idea...But I am not sure whether this is the right way of doing it..please confirm..

[jane2238]

I got the same answer 3.3ml

Jane

[sameh]

i dont think complex calculations will appear in the exam!!!!! its only a minute to answer the question & these calculations take at least 5 mins

[kvr]

thanks jane

kvr

[dinesh123]

yes,the answer is same for me for the first one