In the Minnesota Department of Health set a health risk limit for acetone in groundwater of . Suppose an analytical chemist receives a sample of groundwater with a measured volume of . Calculate the maximum mass in milligrams of acetone which the chemist could measure in this sample and still certify that the groundwater from which it came met Minnesota Department of Health standards. Be sure your answer has the correct number of significant digits.

The question is incomplete, the complete question is;

In the Minnesota Department of Health set a health risk limit for acetone in groundwater of 60.0 μg/L . Suppose an analytical chemist receives a sample of groundwater with a measured volume of 79.0 mL.

Calculate the maximum mass in micrograms of acetone which the chemist could measure in this sample and still certify that the groundwater from which it came met Minnesota Department of Health standards. Round your answer to significant digits.

0.0047mg

Explanation:

From the question, we have the following information;

density of acetone = 60.0 μg/L

Volume of water sample= 79.0 mL

Mass of acetone in water= ?

Solution:

Using:

d = m/v

d= density

m= mass

v=volume

v = 79.0 mL × 1L /1000 mL

v = 0.079 L

Substituting into the given formula:

d = m/v

60.0 μg/L = m/0.079 L

m = 60.0 μg/L × 0.079 L

m = 4.7 μg

maximum mass in milligrams of acetone = 0.0047mg