# Question 8caa0

Feb 2, 2017

To convert

$m m o l \text{/L"->"percentage}$

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

I am converting

$m m o l \text{/L"->g"/100mL}$

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

$= \frac{{10}^{-} 3 \times \text{mol"xxM"g/mol }}{{10}^{3} \times m L}$

$= \frac{{10}^{-} 4 \times M g}{100 m L}$

=Mxx10^-4% (w/v)

$M \text{g/mol"->"molar mass of the substance}$

For 5.4mmol"/L"=5.4xx10^-4M%#