Calculate the molality of 1 litre solution of 93% H2SO4(weight/volume)The density of the soln is 1.84g ml^-1?

molality is defined as the number of moles of solute divided by the mass of the solvent:

m = n/msolvent

So we need to find the number of moles of solute, and the mass of solvent that contains them:

mass of solute in 100 mL of solution = 93g H2S04

divide by the molar mass of H2SO4 to get the number of moles of solute:

93 g H2SO4 / 98.08 = 0.948 moles of H2SO4. This is "n" in the original equation when referring to 100 mL of solution.

Now we need to find the weight of the *solvent* (note this is not the weight of the solution).

The weight of 100 mL of solution is 1.84 g/mL x 100 mL = 184 g of solution.

This is made up of solute + solvent. So 93g solute + Ygrams solvent = 184g solution.

Y = 184 - 93 = 91. So in 100 mL of solution there are 93g of solute, and 91g of solvent.

Now you are ready to calculate the molality:

Molality = n/msolvent = 0.948 / 91 = 0.01 mol/g = 10 mol/kg