Question #37846

Find the molarity and molality of a 15% solution of sulphuric acid whose density is 110g/cc and the molar mass of the sulphuric acid is 98g/mol

Expert's answer

110 g/cc is larger than the density of iridium. Certainly you meant 1.10 g/cc.

To ease the calculations, take 1 liter of the solution. Its mass is

m = V*D = 1000*1.10 = 1100 g = 1.10 kg

The mass of sulfuric acid is

m(acid) = m(solution)*w(acid) = 1100*0.15 = 165 g

m(water) = m(solution) - m(acid) = 935 g

The molarity is defined as the number of moles of the solute per liter of the solution.

The molality is defined as the number of moles of the solute per kilogram of the solvent.

To obtain the molarity and molality of the solution we need the number of moles of acid:

n(H2SO4) = m(H2SO4)/M(H2SO4) = 1.684 mol

Hence we can obtain the molarity and molality:

c(H2SO4) = n(H2SO4)/V(solution) = 1.68 mol/l

μ(H2SO4) = n(H2SO4)/m(water) = 1.684/0.935 = 1.80 mol/kg

To ease the calculations, take 1 liter of the solution. Its mass is

m = V*D = 1000*1.10 = 1100 g = 1.10 kg

The mass of sulfuric acid is

m(acid) = m(solution)*w(acid) = 1100*0.15 = 165 g

m(water) = m(solution) - m(acid) = 935 g

The molarity is defined as the number of moles of the solute per liter of the solution.

The molality is defined as the number of moles of the solute per kilogram of the solvent.

To obtain the molarity and molality of the solution we need the number of moles of acid:

n(H2SO4) = m(H2SO4)/M(H2SO4) = 1.684 mol

Hence we can obtain the molarity and molality:

c(H2SO4) = n(H2SO4)/V(solution) = 1.68 mol/l

μ(H2SO4) = n(H2SO4)/m(water) = 1.684/0.935 = 1.80 mol/kg

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