Question #b844d

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Question #b844d

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Answer 1 · 2017-06-29T00:25:19

$\frac{10 m g N O_{3}}{L i t e r}$ $(10 mgNO_3)/(Liter)$ = $(\frac{0.01 g N O_{3}}{1000 g Soln})$ $((0.01gNO_3)/(1000g"Soln"))$

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

=> $% N O_{3}$ $%NO_3$ = Parts per Hundred = $(\frac{0.01 g N O_{3}}{1000 g Soln})$ $((0.01gNO_3)/(1000g"Soln"))$ x 100% = $1 \times 10^{- 3} % (N O_{3})$ $1xx10^-3%(NO_3)$ or $1 \times 10^{- 3}$ $1xx10^-3$ parts per hundred (pph)

=> Parts per Million $(ppm)$ $("ppm")$ = $(\frac{0.01 g N O_{3}}{1000 g Soln})$ $((0.01gNO_3)/(1000g"Soln"))$ x $10^{6} (ppm)$ $10^6("ppm")$ = 10 ppm

=> Parts per Billion $(ppb)$ $("ppb")$ = $(\frac{0.01 g N O_{3}}{1000 g Soln})$ $((0.01gNO_3)/(1000g"Soln"))$ x $10^{9} (ppb)$ $10^9("ppb")$ = 10000 ppb

=> Molarity(M) = $\frac{moles}{Liter}$ $"moles"/"Liter"$ = $(\frac{0.01 g N O_{3}}{62 \frac{g}{m o l}}) (L^{- 1})$ $((0.01gNO_3)/(62(g)/(mol)))(L^-1)$ = $1.61 \times 10^{- 4} M$ $1.61xx10^-4M$

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Question #b844d

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