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A 500 mL sample of drinking water contains 0.001 mg of mercury. How many ppb is this?

User SkyNT
by
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2 Answers

1 vote

Answer:

2 ppb

Step-by-step explanation:

I got it right in class!

Hope this Helps!! :))

User Boryn
by
7.0k points
4 votes

Answer:

The value is
k = 2 \ ppb

Step-by-step explanation:

From the question we are told that

The volume of the drinking water is
V = 500 \ mL = 500 *10^(-3) \ L = 500 *10^(-6) \ m^3

The mass of mercury it contains
m_c = 0.001 \ mg = 0.001 *10^(-3) \ g = 0.001 *10^(-6)\ kg

Generally the mass of water is mathematically represented as


m = V * \rho

Here
\rho is the density of water with value
\rho = 1000 \ kg/m^3

So


m = 500 *10^(-6 )* 1000

=>
m = 0.5 \ kg

Generally ppb mean part per billon , and 1 billon is 100,000,000

So the part per billon which this represents is mathematically represented as


k = (0.001 *10^(-6))/( 0.5) * 100 000 000

=>
k = 2 \ ppb

User Amir Shirazi
by
7.2k points