The ppm level of F$$^-$$ in a $$500$$ g sample of a tooth paste containing $$0.2$$ g F$$^-$$ is

Parts per million (ppm) for a solute is defined as

$$\text{ppm}=\frac{\text{mass of solute}}{\text{mass of sample}}\times10^{6}$$

Here the mass of fluoride ion in the toothpaste sample is $$0.2\ \text{g}$$ and the total mass of the toothpaste sample is $$500\ \text{g}$$. Since both quantities are already in the same unit, we can substitute them directly:

$$\text{ppm of }F^-=\frac{0.2}{500}\times10^{6}$$

Simplify the fraction first:

$$\frac{0.2}{500}=0.0004=4\times10^{-4}$$

Now multiply by $$10^{6}$$:

$$4\times10^{-4}\times10^{6}=4\times10^{2}=400$$

Therefore the fluoride ion concentration is $$400\ \text{ppm}$$.

Option A which is: $$400$$