Question

The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of concentrated sulfuric acid if the density of the solution is 1.83 g cm-3.

Answer 1

Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution =
`1.83g/cm^3=1.83g/ml`

Molar mass of sulfuric acid (solute) = 98.079 g/mole

98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

Mass of sulfuric acid (solute) = 98.0 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

`\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=(100g)/(1.83g/ml)=54.64ml`

Now we have to calculate the molarity of solution.

`Molarity=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{volume of solution}}=(98.0g* 1000)/(98.079g/mole* 54.64ml)=18.29mole/L`

Now we have to calculate the molality of the solution.

`Molality=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Mass of solvent}}=(98.0g* 1000)/(98.079g/mole* 2g)=499.59mole/Kg`

Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.