Chemistry, 04.03.2020 02:22 bgallman153p71edg

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?

Answers: 3

Show answers

Answers

Answer from: theblondeone2

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

$m=d\times V' = 0.789 g/ml\times 40.0 mL=31.56 g$

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

$m'=d'\times V'' = 1.00 g/ml\times 60.0 mL=60.0 g$

a.)

Moles of ethanol = n= $\frac{31.56 g}{46g/mol}=0.6861 mol$

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

Molarity of the ethanol:

$=\frac{0.6861 mol}{0.100 L}=6.861 M$

6.861 M the molarity of ethanol in this vodka.

b) Mass of ethanol = 31.56 g

Moles of ethanol = n= $\frac{31.56 g}{46g/mol}=0.6861 mol$

Volume of vodka = V = 100 mL

Mass of vodka = m

Density of the water = D = 0.935 g/mL

$M=D\times V=0.935 g/ml\times 100 ml=93.5 g$

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

$\frac{31.56 g}{93.5 g}\times 100=33.75\%$

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

Moles of ethanol = n= $\frac{31.56 g}{46g/mol}=0.6861 mol$

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

Molality of ethanol in vodka :

$m=\frac{0.6861 mol}{0.060 kg}=11.435 m$

11.435 m is the molality of ethanol in this vodka.

Moles of ethanol = $n_1=\frac{31.56 g}{46g/mol}=0.6861 mol$

Moles of water = $n_2=\frac{60.0 g}{18 g/mol}=3.333 mol$

Mole fraction of ethanol = $\chi_1$

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.6861 mol}{0.6861 mol+3.333 mol}$

= 0.1707

Mole fraction of water = $\chi_2$

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{3.3333 mol}{0.6861 mol+3.333 mol}$

= 0.8290

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

$P=\chi_1\times p_1+\chi_2\times p_2$

$P=0.1707\times 45.0torr+0.8290\times 23.8 Torr=27.41 torr$

The vapor pressure of vodka is 27.41 Torr.

Answer from: Quest

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.

Answer from: Quest

As cooking food that has to go in the oven

Another question on Chemistry

Chemistry, 21.06.2019 17:30

Aspirin has a density of 1.40 g/cm3 what is the volume in cubic centimeters of a tablet weighing 320 mg ?

Answers: 1

Answer

Chemistry, 22.06.2019 10:50

How many grams of oxygen gas are contained in a 15 l sample at 1.02 atm and 28°c? show your work.

Answers: 1

Answer

Chemistry, 22.06.2019 16:30

How many moles of sulfuric acid (h2so4) are needed to react completely with 6.8 moles of lithium hydroxide (lioh)? 2lioh + h2so4 → li2so4 + 2h2o a. 3.4 mol h2so4b. 6.8 mol h2so4 c. 10.2 mol h2so4 d. 13.6 mol h2so4

Answers: 3

Answer

Chemistry, 22.06.2019 19:10

Which statement correctly describes the phosphate ion, ? it is composed of one phosphorus atom and four oxygen atoms covalently bonded together, and there is a –3 charge distributed over the entire ion. it is composed of one phosphorus atom and four oxygen atoms covalently bonded together, and there is a –3 charge on the phosphorus atom. it is composed of one phosphorus atom and four oxygen atoms ionically bonded together, and there is a –3 charge distributed over the entire ion. it is composed of one phosphorus atom and four oxygen atoms ionically bonded together, and there is a –3 charge on the phosphorus atom.

Answers: 3

Answer

You know the right answer?

Grey Goose ® vodka has an alcohol content of 40.0 % (v/v). Assuming that vodka is composed of only e...

Questions