Question

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

Answer 1

2 ppb

Step-by-step explanation:

I got it right in class!

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Answer 2

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`