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(a) The given values are:
Volume of solution = 1.00 L
Molarity of CH3CH2OH = 6.86 M CH3CH2OH
Molar mass of CH3CH2OH = 46.07 g/mol
(a) How many grams of ethanol, CH3CH2OH, should you dissolve in water to make 1.00 L of vodka (which is an aqueous solution that is 6.86 M ethanol)?
(b) Using the density of ethanol (0.789 g/mL), calculate the volume of ethanol you need to make 1.00 L of vodka.
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Based on our data, we think this problem is relevant for Professor Johnson's class at GS.