Question

The famous toothpaste Forhans contains $ {\text{0}}{\text{.76}} $ g of sodium per gram of sodium monofluoro orthophosphate ( $ {\text{N}}{{\text{a}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}{\text{F}} $ ) in 100 mL.
How many fluorine atoms are present?
How many fluorine in milligrams is present?

Answer 1

Hint: In the above question, we are asked to find the number of fluorine atoms and weight of fluorine. We are provided with the information of weight of sodium and hence, we can find the number of moles of Na present and since, in the molecule, $ {\text{N}}{{\text{a}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}{\text{F}} $ , 3 moles of Na and 1 mole of F is present and hence, we can find the number of moles of F and after using number of moles formula, we can find the weight of the fluorine.

Formula Used
 $ {\text{n = }}\dfrac{{\text{m}}}{{\text{M}}} $
Where n is the number of moles of gas
m is the given mass
M is the molar mass.

Complete Step by step solution
Let us first find out number of moles of Na present in the molecule.
Molar mass of Na =23
No. of moles of Na = $ \dfrac{{\text{m}}}{{\text{M}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.76}}}}{{{\text{23}}}}{\text{ = 0}}{\text{.03}} $ moles
No. of moles of F = No. of moles of Na/ 3 = $ {\text{0}}{\text{.01}} $ moles
1 mole of a compound contains $ {{\text{N}}_{\text{A}}} $ atoms. Since, $ {{\text{N}}_{\text{A}}} $ is constant quantity which is numerically equal to $ {\text{6}}{\text{.022 \times 1}}{{\text{0}}^{{\text{23}}}} $ .
Hence we can write , $ {\text{0}}{\text{.01}} $ moles of F contains $ {\text{6}}{\text{.022 $\times $1}}{{\text{0}}^{{\text{23}}}}{\text{ $\times$ 0}}{\text{.01 = 6}}{\text{.022 $\times$ 1}}{{\text{0}}^{{\text{21}}}} $ atoms
So, $ {\text{6}}{\text{.022 $\times$ 1}}{{\text{0}}^{{\text{21}}}} $ atoms of fluorine is present.
We know that molar mass of fluorine is 19
Now we find the weight of the fluorine is by the equation:
  $ {\text{n = }}\dfrac{{\text{m}}}{{\text{M}}} $
Rearranging the equation, we get:
 $ {\text{m = n $\times$ M}} $
Now, substituting the values, we get:
 $ {\text{m = n $\times$ M = 0}}{\text{.01 $\times$ 19 = 0}}{\text{.019}} $ g = 19mg
So, the weight of fluorine is 19mg.

Note
1 mole of a substance is numerically equivalent to $ {\text{6}}{\text{.022 \times 1}}{{\text{0}}^{{\text{23}}}} $ particles. These particles can be ion, molecules, atoms etc. So, while the problem we should see with which quantity, it is equal.