Question

Grey Goose ® vodka has an alcohol content of 40.0 % (v/v). Assuming that vodka is composed of only ethanol and water answer the following questions. Note: The molar masses of water and ethanol are 18.0 g and 46.0 g, respectively. The densities of water, ethanol, and this vodka mixture are 1.00 g/mL, 0.789 g/mL, and 0.935 g/mL, respectively

a. Calculate the molarity of ethanol in this vodka, assuming that water is the solvent.
b. Calculate the percent by mass of ethanol % (m/m) in this vodka.
c. Calculate the molality of ethanol in this vodka assuming that water is the solvent.
d. Calculate the mole fractions of ethanol and water in this vodka.
e. Calculate the vapor pressure, in torr, of this vodka at 25.0 oC if the vapor pressures of pure water and ethanol are 23.8 torr and 45.0 torr, respectively?

Answer 1

Step-by-step explanation:

Grey Goose vodka has an alcohol content of 40.0 % (v/v).

Volume of vodka = V = 100 mL

This means that 40.0 mL of alcohol is present 100 mL of vodka.

Volume of ethanol=V' = 40.0 mL

Mass of ethanol = m

Density of the ethanol = d = 0.789 g/mL

`m=d* V' = 0.789 g/ml* 40.0 mL=31.56 g`

Volume of water = V''= 100 ml - 40.0 mL = 60.0 mL

Mass of water = m'

Density of the water = d' = 1.00 g/mL

`m'=d'* V'' = 1.00 g/ml* 60.0 mL=60.0 g`

a.)

Moles of ethanol = n=
`(31.56 g)/(46g/mol)=0.6861 mol`

Volume of vodka = V = 100 mL = 0.100 L ( 1mL=0.001 L)

Molarity of the ethanol:

`=(0.6861 mol)/(0.100 L)=6.861 M`

6.861 M the molarity of ethanol in this vodka.

b) Mass of ethanol = 31.56 g

Moles of ethanol = n=
`(31.56 g)/(46g/mol)=0.6861 mol`

Volume of vodka = V = 100 mL

Mass of vodka = m

Density of the water = D = 0.935 g/mL

`M=D* V=0.935 g/ml* 100 ml=93.5 g`

The percent by mass of ethanol % (m/m):

`(31.56 g)/(93.5 g)* 100=33.75\%`

33.75% is the percent by mass of ethanol % (m/m) in this vodka.

c)

Moles of ethanol = n=
`(31.56 g)/(46g/mol)=0.6861 mol`

Mass of solvent that is water = 60.0 g = 0.060 kg ( 1g = 0.001 kg)

Molality of ethanol in vodka :

`m=(0.6861 mol)/(0.060 kg)=11.435 m`

11.435 m is the molality of ethanol in this vodka.

d)

Moles of ethanol =
`n_1=(31.56 g)/(46g/mol)=0.6861 mol`

Moles of water =
`n_2=(60.0 g)/(18 g/mol)=3.333 mol`

Mole fraction of ethanol =
`\chi_1`

`\chi_1=(n_1)/(n_1+n_2)=(0.6861 mol)/(0.6861 mol+3.333 mol)`

= 0.1707

Mole fraction of water =
`\chi_2`

`\chi_2=(n_2)/(n_1+n_2)=(3.3333 mol)/(0.6861 mol+3.333 mol)`

= 0.8290

e)

The vapor pressure of vodka = P

Mole fraction of ethanol =
`\chi_1=0.1707`

Mole fraction of water =
`\chi_2=0.8290`

The vapor pressures of ethanol =
`p_1=45.0 Torr`

The vapor pressures of pure water =
`p_2=23.8Torr`

`P=\chi_1* p_1+\chi_2* p_2`

`P=0.1707* 45.0torr+0.8290* 23.8 Torr=27.41 torr`

The vapor pressure of vodka is 27.41 Torr.