Medical math?

Question

A sedative solution contains 10mg/ml of a particular drug. What volume of the sedative must be administered to an 18lb dog to achieve a dosage of 2mg/kg?

Answer 1

The dog is 18 lbs which is 8.2 kg (18 / 2.2) (2.2 lbs = 1 kg)

Dose =  8.2 x 2mg = 16.4 mg

If there are 10 mg/ml, then you would need 16.4/10 = 1.64 ml for the final volume.

Answer 2

An 18 pound dog is 8.2 kg.

2 mg / kg x 8.2 kg = 16.4 mg

That would require 1.64 ml.

Answer 3

yup

Answer 4

.82 Ml of solution

18lbs = 8.2 kg

Answer 5

(1) First you want to convert the dog's weight from pounds to kilograms so you can work the rest of the problem.

2.2 lbs = 1 kg, so 18 lbs = 8.18 kg.

(2) Find the dose of medicine needed for the dog.

If the dog needs 2 mg of medicine for each kg, then it would need  16.36 mg of medicine

8.18 kg X 2 mg/kg = 16.36 mg

(3) Find the volume of solution that contains 16.36 mg of the sedative in it.

If the solution contains 10 mg/mL and the dose needed is 16.36 mg, you will need to give a volume of 1.64mL.

16.36 mg / 10 mg/mL = 1.64 mL

Anytime you do medication math make sure that your units cancel out to give you the correct final unit...for example in this problem you were left with mL as the final unit because the milligrams were canceled out when I divided. It's a good way to check yourself after each step, especially when the problem has a lot of steps.

Answer 6

Just a smidge over 1 1/2 teaspoons.

Answer 7

2mg of drug per kg of "dog" is what you want.

First convert 18 lbm to Kg

18 lbs  / 2.2 lbs/kg  = 8.182 kg of dog

10 mg / ml of drug times your desired volume gives the amount (mass) of drug:

10 mg/ml * V ml

To convert to dosage, divide your amount of drug by the mass of the dog, and set it equal to 2 mg/kg (units are in brackets):

(  10 [mg / ml] *  V [ml]  ) / 8.182 [kg]   =  2 [mg/ml]

Solving for V, you get roughly 1.64 mL of solution.

Answer 8

It's really a very simple multiplication/division problem... try doing it yourself first.

Answer 9

1.64 ml?

Answer 10

part of the whole volume.