Question

Express the concentration of a 0.0350 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.a. Number in percentage ....%
b. Number in ppm ........ppm.

Answer 1

Answer :

(a) The mass percentage is 0.0665 %

(b) The concentration in ppm is 665 ppm.

Explanation :

Part (a) :

As we are given that, 0.0350 M aqueous solution of fluoride ion that means 0.0350 moles of fluoride ion present in 1 L of solution.

First we have to calculate the mass of fluoride ion.

$\text{Mass of }F^-=\text{Moles of }F^-\times \text{Molar mass of }F^-$

Molar mass of $F^-$ = 19 g/mole

$\text{Mass of }F^-=0.0350mol\times 19g/mol=0.665g$

Now we have to calculate the mass of solution.

Mass of solution = Density of solution × Volume of solution

Density of solution = 1.00 g/mL

Volume of solution = 1 L = 1000 mL

Mass of solution = 1.00 g/mL × 1000 mL

Mass of solution = 1000 g

Now we have to calculate the mass -percentage.

$\text{Mass of percentage}=\frac{\text{Mass of }F^-}{\text{Mass of solution}}\times 100$

$\text{Mass of percentage}=\frac{0.665g}{1000g}\times 100=0.0665\%$

Thus, the mass percentage is 0.0665 %

Part (b) :

Parts per million (ppm) : It is defined as the mass of a solute present in one million $(10^6)$ parts by the mass of the solution.

Now we have to calculate the concentration in ppm.

$\text{Concentration in ppm}=\frac{\text{Mass of }F^-}{\text{Mass of solution}}\times 10^6$

$\text{Concentration in ppm}=\frac{0.665g}{1000g}\times 10^6=665ppm$

Thus, the concentration in ppm is 665 ppm.