Question

battery acid is 4.22 M aqueous H2SO4 solution, has density of 1.21 g/cm³. What is the molality of H2SO4 ?

Answer 1

This is a 4.22M solution.
1000cm³  of the solution contains 4.22 moles of H2SO4

Molar mass of H2SO4 = 1x2 + 32 + 16x4 = 98

Mass of H2SO4 in the solution = 4.22x98 = 413.56g

Total mass of the solution = 1000x1.21 = 1210g

Mass of water = 1210 - 413.56 = 796.44 g = 0.79644kg

Molality = Moles of solute/ kg of solvent

∴ Molality = 4.22/0.79644 = 5.3mol/kg

Answer 2

5.298 mol kg^-1 (approx.)

Given:

Density of the solution = 1.21 g cm^-3

Molarity of the soln. = 4.22 M

To Calculate:

The Molality of the soln.

Solving:

Let us assume the solution to be 1 liter

Calculating the molecular mass of H2SO4 = 1 g x 2 + 32 g x 1 + 16 g x 4

= 2 + 32 + 64

= 98 g mol^-1

Formula used for calculating Molarity of solution

= Number of moles in solute / Volume of the soln.

Calculating number of moles of solute = Molarity x Volume of the solutions

Substituting the values into this formula we get:

= 4.22 x 1

= 4.22

Density of the solution

= 1.21 g/cm^-3

= 1.21 g/mL

Converting into kg/L:

= 1.21 x 10^3 g/L

= 1.21 Kg/L

Calculating the mass of the solution:

= Density x Volume of the solution

= 1 L x 1.21 kg/L

= 1.21 kg

Calculating the mass of solute:

= Number of moles x Molecular mass

Substituting the values into this formula we get:

= 4.22 x 98

= 413.56 g

Converting into Kilograms we get (By Dividing by 1000):

= 0.41356 kg

Calculating the mass of solvent:

= Mass of solution - Mass of solute

Substituting the values into this formula:

= 1.21 - 0.41356

= 0.79644 kg

Calculating Molality of the solution:

= Number of moles of solute / Mass of solvent in kg

Substituting the values in this formula we get:

= 4.22 mol / 0.79644 kg

= 5.298 mol kg^-1

Therefore, the molality of the solution is 5.298 mol kg^-1 (approx.)