7. (c) Patient weight is 156 lbs, therefore weight in Kg would be 156/2.2 = 70.9 kg

A normal therapeutically recommended dose of drug is 10mg/kg/day, therefore dose in above patient

= 10 x 70.9

= 709 mg

Each ready-infusion-bag contains 250 mg of drug, so number of bags require to fill order would be

= 709/250 = 2.83 @ 3 bags.

8. (b) To solve this kind of problem, we must first find out the amount of drug present in final solution.

Amount of atropine in 1 pint, 1 in 500 soln .

= 480 x 1 /500= 0.96 gm of atropine.

Now, 0.96 gm of drug must be present in 1 teaspoonful of drug solution, therefore we can say

5cc (1 teaspoonful) contains 0.96 gm

240cc solution requires ?

= 240 x 0.96/5 = 46.08 gm of atropine.

9. (d) Gram equivalent weights of solute in 1 liter of solution is defined as normality, therefore 1N solution of sodium bicarbonate will contain 84 gms in 1000 cc. We want to find quantity of sodium bicarbonate in 240cc, 0.1 N solution,

1 N solution contains 84 gm

0.1 N solution contains ?

= 0.1 x 84 = 8.4 gms/1000cc.

240cc solution will contain:

= 240 x 8.4/1000 = 2.016 gm of NaHCO3

10. (d) An amount of sodium chloride presents in 250 cc of 0.9% NaCl,

= 250 x 0.9/100 = 2.25 gm NaCl

Total equivalents Na+ = weight in gm

equivalent wt

= 2.25/58.5 = 0.03846 equivalents

11. (d) The sum of probability of success and failure would be equal to 1 in Null hypothesis and can be expressed by following formula:

p + q = 1, where p = probability of success

q = probability of failure

q = 1 - p

= 1- 0.6

= 0.4

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