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

Answer:

607 ppm

Explanation:

In this case we can start with the ppm formula:

$ppm=\frac{mg~of~solute}{Litters~of~solution}$

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:

$M=\frac{mol}{L}$

In other words:

$0.0320~M=\frac{mol}{1~L}$

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

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

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

$0.0320~mol~F^-\frac{19~g~F^-}{1~mol~F^-}~=~0.607~g$

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

$0.607~g\frac{1000~mg}{1~g}=607~mg$

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

$ppm=\frac{607~mg}{1~L}=~607~ppm$

We will have a concentration of 607 ppm.

I hope it helps!