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The drop of ethanol contains ##3.0xx10^19## molecules of ethanol.
The molar mass of ethanol is ##”46.06844 g/mol”##.
You will first need to determine the moles of ethanol present in the drop by dividing its mass by the molar mass of ethanol. Then calculate the number of molecules of ethanol by multiplying the moles ethanol times ##6.022xx10^23″molecules/mol”##.
##2.3xx10^(-3)cancel(“g CH”_5″OH”)xx(1″mol CH”_5″OH”)/(46.06844cancel(“g CH”_5″OH”))=”0.000050 mol CH”_5″OH”=5.0xx10^(-5) “mol CH”_5″OH”##
##5.0xx10^(-5)”mol CH”_5″OH”xx(6.022xx10^23″molecules CH”_5″OH”)/(1″mol CH”_5″OH”)=3.0xx10^19″molecules CH”_5″OH”##
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