1. what volume of a 5% product is required to prepare 900ml of a 1% solution by

Question

1. what volume of a 5% product is required to prepare 900ml of a 1% solution by adding 3% of a product?

a) 600ml b) 300ml c) 450ml d) none of the above

2. A prescription for 250ml of a cough mixture calls for 2mg of hydrocodone bitartrate per teaspoonful. How many tablets, each containing 5 mg of hdrocodone bitartrate, should be used in preparing the cough mixture?

a) 0.20 b) 0.100 c) 6.6 d) 3.3

3) How many teaspoonfuls in 2,500 units of Humulin 70/30?

a) 5 teaspoonfuls b) 10 teaspoonfuls c) 15 teaspoonfuls d) 25 teaspoonfuls

4) How many grams of dextrose are contained in a pint of D5NS

a) 119 b) 96 C) 48 D) 24

5) A large batch of ointment needs to be compounded. Calculate the number of grams of active ingredient needed to prepare 5kg of 1:10 ointment.

a) 0.5g b) 5,000 g c) 50g d) 500g

6. of the following, what size of bottle will you use to dispense 15 oz of medication?

a) 100mL b) 250mL c) 500 mL d) 1L

7. A prescription for Amoxicillin 75 mg QID for 10 days. How many milliliters do you need to fill this Rx to last the 10 days if you have 250mg/5ml on hand?

a) 20ml b) 40ml c) 60ml d)100ml

8. You receieve a stat order for " Drug X" 35mcg/Kg. The patients reported body weight is 220lb. How much "Drug X" will you dispense?

a) 1.8mg b) 2.8 mg c) 3.5mg d)3,500 mg

8.

Explanation / Answer

1.

5% product is required to prepare 900ml of a 1% solution

3% product is required to prepare (900ml x3/5) = 540 ml of a 1% solution

Ans- d) none of the above

2.

Number of teaspoonful in250ml = 250/5 = 50

2mg of hydrocodone bitartrate per teaspoonful

g of hydrocodone bitartrate in 250 ml= 100g

Number of tablets, each containing 5 mg of hydrocodone bitartrate in 250 ml solution

= 100/5 20 tablet

Ans- a) 20

3)

Humulin70/30 injectable suspension contain 100 units per mL

2,500 units contain2, 500 / 100 = 25ml

5 mL = 1 teaspoon

25ml = 25/5 = 5 teaspoonful

Ans- a) 5 teaspoonful

4) D5NS (5% dextrose in normal saline)

  5% dextrose means the solution contains 5g/100ml of solution

1 pint (pint) = 480 ml

100 ml of solution contains 5g dextrose

480 ml of solution contains480 x( 5g /100) = 24 g dextrose

Ans - D) 24

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