Question

A tube of toothpaste contains 0.76 g of sodium monofluorophosphate(Na2PO3F)in 100 ml.
(a)What mass of florine atoms in mg was present?
(b) How many fluorine in atoms were present?

Answer 1

Explanation:

(a)   Molar mass of $Na_{2}PO_{3}F$ is 691.92 g/mol. And, there is 0.76 g of sodium monofluorophosphate ($Na_{2}PO_{3}F$) 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 = $\frac{19 g/mol}{691.92 g/mol} \times 0.76 g$        

                                           = 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 = $\frac{\text{mass of fluorine}}{\text{molar mass of fluorine}}$

                                         =  $\frac{0.021 g}{19 g/mol}$

                                          = 0.001 mol

Hence, according to mole concept in one mole there are $6.022 \times 10^{23}$ atoms.

So, in 0.001 mole number of fluorine atoms will be calculated as follows.

                  $0.001 mol \times 6.022 \times 10^{23}$ atoms.

                 = $6.022 \times 10^{20}$ atoms

Therefore, there are $6.022 \times 10^{20}$ atoms of fluorine were present.