Respuesta :
Explanation:
(a) Molar mass of [tex]Na_{2}PO_{3}F[/tex] is 691.92 g/mol. And, there is 0.76 g of sodium monofluorophosphate ([tex]Na_{2}PO_{3}F[/tex]) in 100 ml.
Molar mass of fluorine is 19 g/mol.
This means that in 691.92 g/mol there are 19 g/mol of fluorine is present. Hence, mass of fluorine present in 0.76 g is calculated as follows.
Mass of fluorine = [tex]\frac{19 g/mol}{691.92 g/mol} \times 0.76 g[/tex]
= 0.021 g
As, 1 g = 1000 mg. Hence, 0.021 g = 21 mg.
Therefore, mass of florine atoms in present was 21 mg.
(b) As we know that number of moles equal mass divided by molar mass.
Therefore, No. of moles of fluorine = [tex]\frac{\text{mass of fluorine}}{\text{molar mass of fluorine}}[/tex]
= [tex]\frac{0.021 g}{19 g/mol}[/tex]
= 0.001 mol
Hence, according to mole concept in one mole there are [tex]6.022 \times 10^{23}[/tex] atoms.
So, in 0.001 mole number of fluorine atoms will be calculated as follows.
[tex]0.001 mol \times 6.022 \times 10^{23}[/tex] atoms.
= [tex]6.022 \times 10^{20}[/tex] atoms
Therefore, there are [tex]6.022 \times 10^{20}[/tex] atoms of fluorine were present.
