Question

A bottle of wine contains 10.5% ethanol by volume. The density of ethanol (C2H5OH) is 0.789 g/cm3. Calculate the concentration of ethanol in wine in terms of mass percent and molality. mass percent

Answer 1

Answer : The mass percent and molality of ethanol is, 8.28 % and 1.96 mol/kg respectively.

Explanation :

As we are given that 10.5 % ethanol by volume. That means, 10.5 mL of ethanol present in 100 mL of solution.

Density of ethanol =
`0.789g/cm^3`

First we have to calculate the mass of ethanol.

`\text{Mass of ethanol}=\text{Density of ethanol}* \text{Volume of ethanol}`

`\text{Mass of ethanol}=0.789g/cm^3* 10.5cm^3=8.28g`

Now we have to calculate the volume of water.

Volume of water = Volume of solution - Volume of ethanol

Volume of water = (100 - 8.28) cm³

Volume of water = 91.7 cm³

Now we have to calculate the mass of water.

`\text{Mass of water}=\text{Density of water}* \text{Volume of water}`

Density of water = 1.00 g/cm³

`\text{Mass of water}=1.00g/cm^3* 91.72cm^3=91.7g`

Total mass = 8.28g + 91.7g = 99.98 g

Now we have to calculate the mass percent of ethanol.

Mass percent of ethanol =
`\frac{\text{Mass of ethanol}}{\text{Total mass}}* 100=(8.28g)/(99.98g)* 100=8.28\%`

Now we have to calculate the Moles of ethanol.

`\text{Moles of ethanol}=\frac{\text{Given mass ethanol}}{\text{Molar mass ethanol}}=(8.28g)/(46g/mol)=0.18mol`

Now we have to calculate the molality of ethanol.

`\text{Molality}=\frac{\text{Moles of ethanol}}{\text{Mass of water in (kg)}}`

`\text{Molality}=(0.18mol)/(91.7g)=(0.18mol* 1000)/(91.7kg)=1.96mol/kg`

Therefore, the mass percent and molality of ethanol is, 8.28 % and 1.96 mol/kg respectively.