A viral contains radioactive iodine -131 with an activity of 2.0 mCi per milliliter. If the thyroid test requires 3.0 mCi in an "atomic cocktail", how many milliliters are used to prepare the iodine -131 solution ?
2 mCi/mL x ?mL = 3 mCi
Solve for ?mL.
To find out how many milliliters are needed to prepare the iodine-131 solution, we can set up a ratio using the given information. The activity of the iodine-131 solution is 2.0 mCi per milliliter, and the thyroid test requires 3.0 mCi.
Let's assume the volume of the iodine-131 solution we need to find is 'x' milliliters.
The ratio can be set up as follows:
2.0 mCi / 1 mL = 3.0 mCi / x mL
We can cross-multiply and solve for 'x':
2.0 mCi * x mL = 3.0 mCi * 1 mL
2.0x = 3.0
x = 3.0 / 2.0
x = 1.5 mL
Therefore, 1.5 milliliters of the iodine-131 solution are needed to prepare the "atomic cocktail" for the thyroid test.