Calculation of Molality

To calculate the molality of a solution, we need to know the amount of solute (in moles) and the mass of the solvent (in kg). In this case, the solute is 93% H2SO4 and the solvent is water.

Step 1: Calculate the mass of H2SO4

Given that the solution is 93% H2SO4 by weight, we can calculate the mass of H2SO4 in the solution using the density and volume of the solution. The density of the solution is given as 1.84 g/mL and the volume is 1 L.

Mass of solution = density x volume
= 1.84 g/mL x 1000 mL
= 1840 g

Mass of H2SO4 = 93% of mass of solution
= 0.93 x 1840 g
= 1711.2 g

Step 2: Calculate the amount of H2SO4 in moles

To calculate the amount of H2SO4 in moles, we need to know the molar mass of H2SO4. The molar mass of H2SO4 is calculated by adding the atomic masses of its constituent elements.

Molar mass of H2SO4 = (2 x atomic mass of H) + atomic mass of S + (4 x atomic mass of O)
= (2 x 1.008 g/mol) + 32.06 g/mol + (4 x 16.00 g/mol)
= 98.09 g/mol

Amount of H2SO4 in moles = mass of H2SO4 / molar mass of H2SO4
= 1711.2 g / 98.09 g/mol
= 17.45 mol

Step 3: Calculate the mass of the solvent (water)

To calculate the mass of the solvent (water), we subtract the mass of the solute (H2SO4) from the mass of the solution.

Mass of solvent (water) = mass of solution - mass of H2SO4
= 1840 g - 1711.2 g
= 128.8 g

Step 4: Calculate the molality

Molality (m) is defined as the amount of solute (in moles) divided by the mass of the solvent (in kg).

Molality = amount of H2SO4 (in moles) / mass of solvent (in kg)
= 17.45 mol / (128.8 g / 1000 g/kg)
= 135.49 mol/kg

The molality of the 1 L solution of 93% H2SO4 (weight by volume) is 135.49 mol/kg.

104molper kg