# Question b844d

Jun 29, 2017

$\frac{10 m g N {O}_{3}}{L i t e r}$ = $\left(\frac{0.01 g N {O}_{3}}{1000 g \text{Soln}}\right)$

In each of the following, note how the ratio (in common units) is multiplied by the factor needed; that is,
=> pph = ratio x 100 => %
=> ppm = ratio x 1-million
=> ppb = ratio x 1-billion

=> %NO_3 = Parts per Hundred = $\left(\frac{0.01 g N {O}_{3}}{1000 g \text{Soln}}\right)$ x 100% = 1xx10^-3%(NO_3)# or $1 \times {10}^{-} 3$ parts per hundred (pph)

=> Parts per Million $\left(\text{ppm}\right)$ = $\left(\frac{0.01 g N {O}_{3}}{1000 g \text{Soln}}\right)$ x ${10}^{6} \left(\text{ppm}\right)$ = 10 ppm

=> Parts per Billion $\left(\text{ppb}\right)$ = $\left(\frac{0.01 g N {O}_{3}}{1000 g \text{Soln}}\right)$ x ${10}^{9} \left(\text{ppb}\right)$ = 10000 ppb

=> Molarity(M) = $\text{moles"/"Liter}$ = $\left(\frac{0.01 g N {O}_{3}}{62 \frac{g}{m o l}}\right) \left({L}^{-} 1\right)$ = $1.61 \times {10}^{-} 4 M$