a solution of h2so4 with a molal concentration of 6.80 has a density of 1.318 g/

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Assumption: Volume of solution is 1 L (1000 mL) From the equation of molality = (mass of solute / molar mass solute) / kg of solvent, the mass of solute can be solve as: mass of solute(H2SO4) = molality x kg of solvent x molar mass of solute molality = 6.80 molal molar mass of H2SO4 = 98.08 g/mol mass of solvent = mass of solution - mass of solute but: mass of solution = density of solution x volume of solution mass of solution = 1.318 g/mL x 1000 mL = 1318 g mass of solvent = 1318 g - mass of solute ====> divide by 1000 to make the resulting answer in kg so that it will be consistent with the equation of molality. kg of solvent = (1318 - mass of solute) / 1000 ====> equation 1 Going back to the equation of molality: mass of solute(H2SO4) = molality x kg of solvent x molar mass of solute mass of solute (H2SO4) = (6.80 mol/kg) x (kg of solvent) x 98.08 g/mol ===> equation 2 Substitute equation 1 in 2 and Let x be the mass of solute (H2SO4) x = (6.80 mol/kg) x [(1318 - x) / 1000] x 98.08 g/mol ====> solve for x x = mass of H2SO4 = 527.33 g Molarity = (mass of H2SO4 / molar mass of H2SO4) / L of solution Molarity = (527.33 g / 98.08 g/mol) / 1 L Molarity = 5.38 molar =====> answer

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