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a 1. How many milliliters of a 3% hydrogen peroxide solution would be needed to make 120 ml of 1% hydrogen peroxide solution? How many milliliters of water should be added? [40ml & 80ml] 2. The strength of real acetic acid is 33% (W/V). Using this, prepare 200ml of a solution of acetic acid containing 4% (W/V) of real acetic acid. [24.2ml] 3. Prepare 200 ml of a solution of ammonia containing 4% by weight of ammonia. The strong solution of ammonia contains 32.5% of ammonia W/W. [24.62gm] 4. Prepare 400ml of a 5% solution and label with direction for preparing 2 liter quantities of a 1 in 2000 solution. [20gm & 20ml up to 2 liter]
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How many grams of zinc oxide are needed to make 240 grams of a 4% (w/w) zinc oxide ointment? Answer: IF 4 g ZnO
THEN X g ZnO =
100 g Oint
240 g Oint
V 100 X = 960 V X = 9.6 g of zinc oxide
(answer)
How many milliliters of a 5% (w/v) boric acid solution can be made from 20 grams of boric acid? Answer:
(answer) How many milliliters of paraldehyde are needed to make 120 ml of a 10% (v/v) paraldehyde solution? Answer: 12ml
How many grams of ephedrine sulfate are needed to make 120 ml of a 2% (w/v) ephedrine sulfate solution? Answer: 1gm dissolved in 100ml = 1% w/v solution 2gm dissolved in 100ml = 2% w/v solution For 120ml solution the quantity of ephedrine sulfate required2x120
=
2.4 gm of ephedrine sulfate needed. 100 (answer) Prepare 500 ml of a 2 in 4000 solution of potassium permanganate. Answer: 50mg
ml
? KMnO4 Distilled water qsad
0.02% 240
How many grams of potassium permanganate (KMnO4) are needed to compound the above prescription? Answer: 0.048 grams Prepare 150ml of 4% (W/V) potassium permanganate and label the directions for preparing 500ml of 1 in 2500 solution.
a If 5 g of a chemical is dissolved in enough water to make the preparation measure one liter, what is the percentage strength of the solution? How many milliliters of a 0.02% W/V solution can be made from 2.5 g of a chemical? Ź Normal saline solution contains 0.9% W/V NaCl. How many grams of sodium chloride should be used to make 1.5 liters of normal saline? Ź How many grams of strong silver protein (SSP) are required to make 250 ml of a 0.25% (w/v) solution?
Overage means excess. Usually companies use certain percent (within the specification limit) excess of active ingredient in their preparation. Veximco uses raw material of paracetamol in which purity of paracetamol is 94% for manufacturing Napa tablet (500mg paracetamol). The batch size of Napa is 50,000 Tablets. Calculate the amount of Raw Materials of Paracetamol to produce a batch of NAPA. If Veximco uses 3% overage then calculate the amount of RM of paracetamol to manufacture a batch of NAPA.
: !a :" !!# " The parts-per notation is used in some areas of science and engineering because it does not require conversion from weights or volumes to more chemically relevant units such as normality or molarity. It describes the amount of one substance in another. It is the ratio of the amount of the substance of interest to the amount of that substance plus the amount of the substance it is in. denotes the amount of a given substance in a total amount of 100 regardless of the units of measure as long as they are the same. e.g. 1 gram per 100 gram. 1 part in 102.
denotes the amount of a given substance in a total amount of 1000 regardless of the units of measure as long as they are the same. e.g. 1 milligram per gram, or 1 gram per kilogram. 1 part in 103. $$denotes the amount of a given substance in a total amount of 1,000,000 regardless of the units of measure used as long as they are the same. e.g. 1 milligram per kilogram. 1 part in 106. Calculate the quantity of Sodium Fluoride (NaF) required for the preparation of 500ml of 2 ppm NaF aqueous solution. Answer- 0.001 gm
%&%!# " 'ë a a ( The word posology derived from Greek word µposos¶, meaning how much and µlogos¶, means science. That means it is a branch of medical science which deals with doses or quantity of drugs which can be administered to produce the required pharmacological actions. The term dose refers to the amount of medication that a patient must take at one time to produce the optimum therapeutic effect.
) #* According to age: a) Young¶s formula: Child's dose =
Child's age in years Child's age in years + 12 years
X Adult dose
The above formula is used for calculating the doses for children under 12 years of age. Problem: The adult maintenance dose of a drug is 325 mg. Calculate the dosage for a 3-year-old child. Answer: 65mg
b) Dilling¶s formula: Child's dose =
Child's age in years 20
X Adult dose
Vecause of easy and quicker calculations, Dilling¶s formula is considered better. Problem: If the adult dose is 60mg and the age of the child is 6 years, what will be child¶s dose? Answer: 18mg c) Fried's rule: Child's dose =
Child's age in months 150 months
X Adult dose
Problem: The adult dose of an antihistamine is 50 mg. Calculate the dosage for a 2½ year-old child. Usually used for infant¶s dose calculation.
Answer: 10mg
##* According to body weight: Clark's Rule: Child's dose =
Child's weight in lbs. 150 lbs
X Adult dose
Problem: The adult dose of Doxycycline is 100 mg. Calculate the dosage for a child weighing 50 pounds. Answer: 33.3 mg ###* According to body surface area: Child's dose =
Surface area of child Surface area of adult
X Adult dose
The average body surface area for an adult = 1.73m2. Hence,
Child's dose =
Surface area of child 1.73m2
X Adult dose
Problem: Calculate the dosage for a child that has a body surface area of 0.57 m2, when the adult dose of a drug is 50 mg. Answer: 16.5 mg " ( In chemistry, normality is a measure of concentration, it is equal to the number of gram equivalents of a solute per liter of solution. It is denoted by N. Normality =
Gram equivalents Liters solution
Gram equivalents =
Atomic weight or molecular weight Valence
Problem: 5.30 gm of Na2CO3 was dissolved in water and the volume made to 100ml. Calculate the normality of the solution. Answer: Molecular weight of Na2CO3= 106 The equivalent weight of Na2CO3 is half the molecule weight, i.e., 106/2= 53 100 ml solution contains 5.30 gm Na2CO3 Therefore, 1000 ml solution contains (5.30x10) gm = 53 gm Normality of the solution is 1
( It defined as the presence of number of moles of solute in 1000ml or 1 litre of solution. It is denoted by M. Molarity =
Moles solute Liters solution
Problem: 58.44 grams of NaCl dissolved in exactly 2 L of solution. What would be the molarity of the solution? Answer: Molecular weight of NaCl is 58.44 58.44 grams 58.44 grams/mol
=1mol
Then, dividing 1.00 mol by 2.00 L gives 0.500 mol/L (or 0.500 M). Problem: Calculate the molarity of 25.0 grams of KVr dissolved in 750.0 mL. Molecular weight of KVr is 119 . Answer: 0.28 M ( Molality denotes the number of moles of a given substance per kilogram of solvent. (not solution) It is denoted by m. Molarity =
Moles solute Kg Solvent
