Question

Express the concentration of a 0.0320 M aqueous solution of fluoride, F−, in mass percentage and in parts per million (ppm). Assume the density of the solution is 1.00 g/mL.

Answer 1

607 ppm

Step-by-step explanation:

In this case we can start with the ppm formula:

`ppm=(mg~of~solute)/(Litters~of~solution)`

If we have a solution of 0.0320 M, we can say that in 1 L we have 0.032 mol of
`F^-`, because the molarity formula is:

`M=(mol)/(L)`

In other words:

`0.0320~M=(mol)/(1~L)`

`mol=0.032~M*1~L=0.032~mol`

`1~L~of~Solution=0.0320~mol~of~solute`

If we use the atomic mass of
`F` (19 g/mol) we can convert from mol to g:

`0.0320~mol~F^-(19~g~F^-)/(1~mol~F^-)~=~0.607~g`

Now we can convert from g to mg (1 g= 1000 mg), so:

`0.607~g(1000~mg)/(1~g)=607~mg`

Finally we can divide by 1 L to find the ppm:

`ppm=(607~mg)/(1~L)=~607~ppm`

We will have a concentration of 607 ppm.

I hope it helps!