Question 1 a) The recommended daily dosage of pseudoephedrine hydrochloride for a child is 0.05 g/kg...

Question 1

a) The recommended daily dosage of pseudoephedrine hydrochloride for a child is 0.05 g/kg in three divided doses. You have a 15% w/v solution of pseudoephedrine hydrochloride. How many mL of this cold remedy should a 16 kg child be given per dose?

b) Calculate the molarity (M) of a 10% aqueous solution of MgCl2 (molecular weight: 94.

c) You receive a script for Vitamin A solution 3500 units per day. There is a commercially available product containing 3000 units/g. How many drops would you advise the patient to give when the dropper delivers a volume of 2 mL in 25 drops? (Vitamin A product wt/mL 0.9 g)

d) You received a prescription for Codeine linctus 5 mg/mL 100 mL. The market product is currently unavailable so you will have to extemporaneously prepare this. Using the following formula taken from APF 19, work out the amounts to make 100 mL.

            Codeine Linctus APF 19 Formula:

Codeine phosphate                                           25 mg

Purified water                                                   0.5 mL

Glycerol                                                               1 mL

Methyl hydroxybenzoate solution                   0.05 mL

Syrup                                                           to      5 mL

*Please show all working

Homework Answers

Answer #1

Que1. Given , recommended dose for child is 0.5 g /kg per day

Available solution is 15%(w/v),child weight is 16kg

Dose for the child is

If, 0.5g = 1kg

16kg = X g

X= 16×0.5= 8 g

X= 8g /day

Devided in 3 doses in a day so per dose= 8÷ 3

= 2.6666 g per dose

So, one dose is 2.666g

Given, 15% solution means 15g of medication in 100 ml volume of Normal saline or solvent.

If, 15g= 100ml

2.666g = Y ml

Y= (2.666÷15)×100= 17.7733 ml

Y= 17.77ml approx 17.8 ml

So, 17.8ml needed per dose for child.

Que2.Given ,10% of mgCl2 means 10grams of magnesium chloride in 100ml of solvent or normal saline.

Given molar mass=94  

By formula

n( no. Of moles)= mass (in grams )÷ M (molar mass)

= 10g× 94= 940 moles

n = 940moles

If 940moles of solute in 100ml solvent

100ml = 940moles

1000ml =Xmoles

X= (1000÷100)× 940 = 9400moles

X= 9400moles in 1L

So, Molarity = number of moles of solute÷ volume (inL)

M=9400moles ÷1 L

M= 9400


Que3. Given, on script 3500 unit

Available = 3000unit/gram

Available = 0.9g/ml

2ml = 25DROPS

if , 1gram =3000unit

Xgram= 3500unit

X= 3500÷3000= 1.166gram

X= 1.166 grams.

Now if, 0.9g=1ml

1.166g = Yml

Y= 1.166÷0.9 =1.2962 ml

Y= 1.2962ml

If,2ml = 25drops

1.2962ml= Zdrops

Z= (1.2962÷2)× 25= 16.2025 drops

Z= 16.2025 or approx 16 drops

Z=16 drops

16 drops in 1.2962ml needed to administer

Que D. As extemporaneous preparation is for a specific its for codein amount of codein phosphate is 25mg in 5ml syrup .

If,25mg of codein phosphate is there in 5ml

25mg = 5ml

100ml = X mg

, X=(100÷5)× 25

X= 500mg

So we will give the patient 100ml With 500 dose of codein phosphate.

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