Question

How many moles, kmols in: 100 g of CO2, 1 litre of ethyl alcohol of density 0.789 g/cm3 and a) 1.5m3 of O2 at 25°C and 1 atm. b) A mixture of water and ethyl alcohol is made up of 1 litre of water and 2.5 litre of alcohol. Calculate the mass fraction and mol fraction for water and alcohol Density of water 1000 kg/m Density of alcohol 789 kg/m

Answer 1

Step-by-step explanation:

1) Mass of carbon dioxide = 100 g

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide =
`(100 g)/(44 g/mol)=2.273 moles`

1 mol = 0.001 kmol

2.273 moles= 2.273 × 0.001 kmol =
`2.273* 10^(-3) kmol`

2) 1 liter of ethyl alcohol of density
`0.789 g/cm^3`

Volume of ethyl alcohol ,V= 1 L = 1000 mL

Density of ethyl alcohol =d =
`0.789 g/cm^3`

`1 cm^3=1 mL`

Mass of ethyl alcohol = m

`m=d* V=0.789 g/cm^3* 1000 mL=789 g`

Molar mass of ethyl alcohol = 46 g/mol

Moles of ethyl alcohol =
`(789 g)/(46 g/mol)=17.152 mol`

`17.152 mol=17.152* 0.001 kmol=1.7152* 10^(-4) kmol`

3) Volume of oxygen gas,V =
`1.5 m^3=1500 L`

`1 m^3= 1000 L`

Temperature of the gas = T= 25°C = 298.15 K

Pressure of the gas ,P= 1 atm

Moles of oxygen gas = n

`PV=nRT`

`n=(RT)/(PV)=(0.0821 atm L/mol K* 298.15 K)/(1 atm* 1500 L)=0.01632 mol`

0.01632 mol = 0.01632 × 0.001 kmol=
`1.632* 10^(-5) kmol`

4) Volume water in mixture = 1 L

Density of water =
`1000 kg/m^3=(1,000,000 g)/(1000 L)=1000 g/L`

Mass of water =
`1000 g/L* 1 L = 1000 g`

Volume of alcohol = 2.5 L

Density of alcohol =
`789 kg/m^3=(789000 g)/(1000 L)=789 g/L`

Mass of alcohol =
`789 g/L* 2.5 L = 1972.5 g`

Mass of mixture = 1000 g + 1972.5 g = 2972.5 g

Mass percentage of water :

`(1000 g)/(2972.5 g)* 100=33.64\%`

Mass percentage of alcohol :

`(1972.5 g)/(2972.5 g)* 100=66.36\%`

Moles of water :

`n_1=(1000 g)/(18 g/mol)=55.55 mol`

Moles of alcohol =

`n_2=(1972.5 g)/(46 g/mol)=42.88 mol`

Mole fraction of water :

`\chi_1=(n_1)/(n_1+n_2)=(55.55 mol)/(55.55 mol+42.88 mol)=0.5644`

Mole fraction of alcohol :

`\chi_2=(n_2)/(n_1+n_2)=(42.88 mol)/(55.55 mol+42.88 mol)=0.4356`