Respuesta :
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.
[tex]\text{Mass of }F^-=\text{Moles of }F^-\times \text{Molar mass of }F^-[/tex]
Molar mass of [tex]F^-[/tex] = 19 g/mole
[tex]\text{Mass of }F^-=0.0350mol\times 19g/mol=0.665g[/tex]
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.
[tex]\text{Mass of percentage}=\frac{\text{Mass of }F^-}{\text{Mass of solution}}\times 100[/tex]
[tex]\text{Mass of percentage}=\frac{0.665g}{1000g}\times 100=0.0665\%[/tex]
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 [tex](10^6)[/tex] parts by the mass of the solution.
Now we have to calculate the concentration in ppm.
[tex]\text{Concentration in ppm}=\frac{\text{Mass of }F^-}{\text{Mass of solution}}\times 10^6[/tex]
[tex]\text{Concentration in ppm}=\frac{0.665g}{1000g}\times 10^6=665ppm[/tex]
Thus, the concentration in ppm is 665 ppm.
