Question

Calculate the molarity and molality of 20 percent aqueous ethanol solution by volume (density of solution = 0.960 g per cm^3)

Answer 1

20% aqueous ethanol solution by volume means 20 ml/ ethanol in 80 ml/ water.
Density of the solution as 0.960 g/cm3 or 0.960 g/ml
Now,
Mass of solution = density x volume
  = 0.960 x 100
  = 96 g
Mass of 80 ml water = density of water x volume of water
  = 1g/ml x 80 ml = 80 g
Mass of ethanol in solution = mass of solution - mass of water
  = 96-80 = 16 g
Molar mass of ethanol,(C2H5OH) = 12 x 2 + 5x 1 + 16 + 1
  = 46 g
Number of moles of ethanol = 16/46 = 0.348
Molarity = number of moles/volume in litres
  = 0.348/0.1
  = 3.48 M

Molality = number of moles of solute/ mass of solvent in Kg
  = 0.348/0.08
  = 4.35 m

Answer 2

Answer : The molarity and molality of solution is, 3.425 mole/L and 4.269 mole/Kg respectively.

Solution : Given,

Density of solution = $0.960g/cm^3=0.960g/ml$    $(1ml=1cm^3)$

Density of ethanol (solute)= $0.789g/cm^3=0.789g/ml$

Molar mass of ethanol (solute) = 46.07 g/mole

20 % aqueous ethanol solution by volume means that the 20 ml of ethanol is present in the 100 ml of solution.

So, the volume of solution = 100 ml

Volume of ethanol (solute) = 20 ml

First we have to calculate the mass of solution.

$\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}$

$\text{Mass of solution}=0.960g/ml\times 100ml=96g$

Now we have to calculate the mass of solute, ethanol.

$\text{Mass of solute}=\text{Density of solute}\times \text{Volume of solute}$

$\text{Mass of solute}=0.789g/ml\times 20ml=15.78g$

Now we have to calculate the mass of solvent.

$\text{Mass of solvent}=\text{Mass of solution}-\text{Mass of solute}=96-15.78=80.22g$

Now we have to calculate the molarity of solution.

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution in ml}}=\frac{15.78g\times 1000}{46.07g/mole\times 100ml}=3.425mole/L$

Now we have to calculate the molality of solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent in g}}=\frac{15.78g\times 1000}{46.07g/mole\times 80.22g}=4.269mole/Kg$

Therefore, the molarity and molality of solution is, 3.425 mole/L and 4.269 mole/Kg respectively.