Question

What is the concentration in ppm of selenium if 1.3 mg is found in 2500 kg of soil?

Answer 1

$\begin{aligned} mass\ of\ selenium &= 1.3mg = 1.3× 10^{-3}g\\ mass\ of\ soil &= 2500kg = 2.5×10^6g\\ \\ ppm\ of\ selenium &= \dfrac{mass\ of\ selenium(g)}{mass\ of\ soil(g)}× 10^6\\ \\ &= \dfrac{1.3× 10^{-3}g}{2.5×10^6g}× 10^6\\ \\ &= 0.52 × 10^{-9} × 10^6\\ &= 5.2× 10^{-4}ppm\\ &=0.00052ppm \end{aligned}$ $\therefore$ The concentration in ppm of selenium is 0.00052ppm