Question

What mass of sodium arsenate would be present in a 1.20 L sample of drinking water that just meets the standard?

Answer 1

Step-by-step explanation:

It is known that the density of water is 1 kg/L which is also equal to 1000 g/L. And, relation between mass, density and volume is as follows.

Density =
`(mass)/(volume)`

or, mass = Density × volume

Hence, putting the given values into the above formula as follows.

mass = Density × volume

= 1000 g/L × 1.20 L

= 1200 g

Also, we are given that concentration of As = 10 ppb =
`(10)/(10^(9))`

Mass of As = concentration x mass of water

=
`(10)/(10^(9)) * 1200`

=
`1.2 x 10^(-5) g`

Therefore, moles of
`Na_(3)AsO_(4)` are as follows.

moles of As =
`\frac{mass}{\text{molar mass of As}}`

=
`(1.2 x 10^(-5))/(74.92)`

=
`1.601 * 10^(-7)` mol

Hence, calculate the mass of
`Na_(3)AsO_(4)` as follows.

Mass = moles x molar mass of
`Na_(3)AsO_(4)`

=
`1.601 * 10^(-7) * 207.89`

=
`332.83 * 10^(-7)`

=
`3.32 * 10^(-5) g`

Thus, we can conclude that mass of given sodium arsenate is
`3.32 * 10^(-5) g`.