Question #8caa0

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Question #8caa0

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Answer 1 · 2017-02-02T11:15:23

To convert

$m m o l /L \to percentage$ $mmol"/L"->"percentage"$

I assume conversion to $w /v$ $w"/v"$ percentage is wanted here. Since the density of the mixture is not given.

I am converting

$m m o l /L \to g /100mL$ $mmol"/L"->g"/100mL"$

So
$m m o l /L = \frac{10^{- 3} \times mol}{10^{3} \times m L}$ $mmol"/L"=(10^-3xx"mol")/(10^3xxmL)$
$= \frac{10^{- 3} \times mol \times molar mass}{10^{3} \times m L}$ $=(10^-3xx"mol"xx"molar mass")/(10^3xxmL)$

$= \frac{10^{- 3} \times mol \times M g/mol}{10^{3} \times m L}$ $=(10^-3xx"mol"xxM"g/mol ")/(10^3xxmL)$

$= \frac{10^{- 4} \times M g}{100 m L}$ $=(10^-4xxMg)/(100mL)$

$= M \times 10^{- 4} %$ $=Mxx10^-4%$ (w/v)

$M g/mol \to molar mass of the substance$ $M"g/mol"->"molar mass of the substance"$

For $5.4 m m o l /L = 5.4 \times 10^{- 4} M %$ $5.4mmol"/L"=5.4xx10^-4M%$

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