Question

A sample toothpaste weighing 500g, on analysis was found to contain 0.2g of fluorine. The concentration of fluorine in ppm is
A.$4 \times {10^3}$
B.$4 \times {10^2}$
C.$4 \times 10$
D.$2 \times {10^2}$

Answer 1

Hint: We have to know that the fluorine is a chemical element, atomic number is $9$. It is used in toothpaste as an ingredient for teeth strength and suppress the acid production and also suppress the activity of plaque.
We have to remember that the ppm is an abbreviation for parts per million. And also expressed as milligram of substance per liter$(mg/l)$. This measurement is used in the mass of chemical or contaminate per unit volume of chemicals or water.

Complete step by step solution:
Given data contains,
Mass of toothpaste is \[500\]g (or $0.5$kg)
The mass fluorine content in the toothpaste is \[0.2g\]($200mg$).
So the concentration per $kg$ is ($\dfrac{{0.2}}{{500}} \times 1000g$) equal to $0.4g$.
We need to convert the concentration milligram to find the level of fluoride ion:
So multiply by mass volume by 1000 to get milligram, the concentration of fluoride ion,
$\dfrac{{mg}}{{kg}} = 0.4 \times 1000mg$
On simplifying we get,
$ \Rightarrow \dfrac{{mg}}{{kg}} = 400mg$
Hence, the level of fluorine in parts per million, so $400ppm$.
A sample toothpaste weighing \[500\]g, on analysis was found to contain \[0.2g\left( {200mg} \right)\]of fluorine. The concentration of fluorine in 1kg of toothpaste is $200 \times 2 = 400ppm$.
The correct option is (2) $4 \times {10^2}$.

Note:Now we discuss another method to calculate to find the concentration of fluoride in ppm in the toothpaste as,
The concentration of fluorine in ppm = number of parts of the component (fluorine mass) / total number of parts of all component (sample: toothpaste)
$0.2g/500g$=$0.0004g$
Multiply the above value by one million to get the value in ppm
$(0.0004 \times 1000000)$= $400ppm$.
Parts per million (or ppm) is important when the particular solute is present in trace quantities. It is an easy way to express the concentration of solute in that case.