Question

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.

Answer 1

m = 4.7 μg

Step-by-step explanation:

Given data:

density of acetone = 60.0 μg/L

Volume = 79.0 mL

Mass = ?

Solution:

Formula:

d = m/v

v = 79.0 mL × 1L /1000 mL

v = 0.079 L

Now we will put the values on formula:

d = m/v

60.0 μg/L = m/0.079 L

m = 60.0 μg/L × 0.079 L

m = 4.7 μg

So health risk limit for acetone = 4.7 μg