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Answer 1 · 2017-02-02T11:15:23

To convert

#mmol"/L"->"percentage"#

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

I am converting

#mmol"/L"->g"/100mL"#

So
#mmol"/L"=(10^-3xx"mol")/(10^3xxmL)#
#=(10^-3xx"mol"xx"molar mass")/(10^3xxmL)#

#=(10^-3xx"mol"xxM"g/mol ")/(10^3xxmL)#

#=(10^-4xxMg)/(100mL)#

#=Mxx10^-4%# (w/v)

#M"g/mol"->"molar mass of the substance"#

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