Question

4.Calculate the molarity of the ethanol in the undiluted vodka sample. As in the lab procedure, the original (undiluted) vodka was diluted 1/40

5.Calculate the mass of ethanol (CH33​CH22​OH) in a litre of vodka.

6.Given that the density of ethanol (CH33​CH22​OH) is 0.789 g⋅mL−1−1, convert the mass of ethanol into a volume of ethanol in a litre of vodka.

7.When 2 miscible liquids are mixed, very seldom does the final volume equal the combined volumes of the individual liquids. The 2 liquids solvate each other and the final volume is almost always less. This is called non-ideal behaviour. Since ethanol and water are infinitely miscible, vodka is an example of a non-ideal solution and the volume of ethanol is determined by multiplying by
976 mL/1000mL.
What is the corrected volume of ethanol in 1 liter of vodka?

8.Report the % ethanol (v/v) in the vodka sample.

% ethanol (v/v) = (volume of ethanol in mL / 1000 mL) x 100 %

Answer 1

Ans. 4:

Vodka is a solution that contains 40% ethanol and 60% water .

ie in 1L of vodka there is 400ml of ethanol and 600 ml of water.

Also , Molar mass of ethanol [ CH3CH2OH] is 46 .

now Molarity is given by , M = no of moles of solute / volume of solution in L .........(1)

here volume of solution ie vodka is 1L .

Now calculation of no of moles of solute , n = given mass / molar mass

given mass = vol * density = 400 * 0.789 = 315.6 gm => 315.6 gm /976 ml ie 323.3 gm/L which means given mass = 323.3 gm

n = 323.3 / 46 = 7.02 moles

put value of n in equation (1)

=> molarity = 7.02 moles / 1L

Ans 5.

Mass of ethanol in one litre of vodka

As vodka is a solution that contains 40% ethanol and 60% water .

ie in 1L of vodka there is 400ml of ethanol and 600 ml of water. Which implies that 40% of ethanol is used in 1L of vodka ,

   => 46 x 40 /100 = 18.4 g