Question

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Automobile batteries contain aqueous sulfuric acid, H_2SO_4. A 3.75 M solution of aqueous sulfuric acid has a density of 1.230 g/mL. What is the mass percent H_2SO_04 in this solution? What is the molality of this solution?. When 0.500 g of an unknown nonelectrolyte is dissolved in 25.00 g of camphor, the freezing point of the solution is 8.68 degree C lower than the freezing point of pure camphor. Determine the molar mass of the unknown. For camphor, K_f = 40.0 degree C/m.

Answer 1

Ans. 1. Given, strength of sulfuric acid = 3.75 M

                        Density = 1.230 g/mL

Now,

3.75 M mean 3.75 mol of sulfuric acid in 1.0 L os solution.

            Mass of H2SO4 in solution = Moles x molar mass

                                                            = 3.75 mol x (98.07948 g/mol)

                                                            = 367.79805 g

            Mass of 1.0 L solution = Volume x density

                                                = 1.0 L x (1.230 g/mL)

                                                = 1230.0 g

                                                = 1.230 kg

#1. A. Mass % of H2SO4 = (mass of H2SO4/ Mass of solution) x 100

                                    = (367.79805 g / 1230.0 g)

                                    = 29.90 %      

#1. B. Molality of H2SO4 = moles of H2SO4/ mass of solvent (i.e. water) in kg

                                    = 3.75 mol / 1.230 kg

                                    = 3.05 mol/kg

                                    = 3.05 m

Ans. 2. Ans. Freezing point depression, dT_f is given by-

            dT_f = i K_f m                - equation 1

            where, i = Van’t Hoff factor = 1 for non-dissociating solutes

                        K_f = molal freezing point depression constant of the solvent

                        m = molality of the solution

                        dT_f = Freezing point of pure solvent – Freezing point of solution

                                    = depression in freezing point.

Let the molar mass of the compound be Y g/mol.

So, number of moles of the compound = mass/ molar mass

                                                = 0.500 g/ (Y g/mol)

                                                = (0.50 / Y) mol

Molality of solution = Moles of solute / Mass of solvent in kg

                                    = (0.50 / Y) mol / 0.025 kg                  ; [1 kg = 1000 g]

                                    = (20/ Y) mol/ kg

                                    = (20/ Y) m               

Given,

            Depression in freezing point, dT_f = 8.68⁰C

            K_f = 40.0⁰C/ m

Putting the values in equation 1-

            8.68⁰C = 1 x (40.0⁰C/ m) x (20/ Y) m

            Or, 8.68⁰C = 1 x (40.0⁰C/ m) x (20/ Y) m

            Or, 8.68⁰C = 800⁰C/ Y

            Or, Y = 800⁰C / 8.68⁰C = 92.165

Hence, molar mass of the compound = Y g/mol = 92.165 g/mol