Calculate the molarity and molality of 93% H2SO4(w/v). The density of the solution is 1.84 g/mL.

Density of solution = 1.84 g mL-1 = massvolume=1.84 g1000 mL

So mass of solution in 1 L = density × volume

= 1.84 g mL​-1 × 1000 mL

= 1840 g

Mass % of H2SO4 (w/v) = 93 %
i.e. mass of solute H2SO4 in 100 mL = 93 g
So mass of H2SO4 in 1 L (1000 mL)= 930 g

Thus mass of solvent = mass of solution - mass of solute

= 1840 - 930 g

= 910 g = 0.910 kg

Hence Molality = mass of solutemolar mass of solute × mass of solvent in kg

= 93098 × 0.910

= 10.43 m


Please ask molarity of the solution as a seperate question.We are not supposed to answer multiple questions.im really sorry for the inconvenience.I understand that both are a part of a question.But try posting the second part as a seperate question.

Hope this may help you.