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------------------------------------------------ Determine the molality of an aqueous solution that is 2.45 M in methanol if its density is 0.976 g/mL. (Molar mass of methanol = 32.04 g/mol)

Perchloric acid, [tex]HClO_4[/tex], can be purchased at a concentration of 9.20 M. If the density of the solution is 1.54 g/mL, what is its percent by mass? (Molar mass of [tex]HClO_4[/tex] = 100.5 g/mol)

To answer this, we first need to establish what molality and molarity are. Molality is the amount of solute (in moles) divided by the mass of the solvent (in kilograms), while molarity is the amount of solute (in moles) divided by the volume of the solution (in liters).

Given that we know the molarity (M) and the density (D), we can calculate the molality (m) using the formula: m = M / ((1/D) - (M / Mm)), where Mm is the Molar mass. Substituting the given values for methanol: m = 2.45 M / ((1 / 0.976 g/mL) - (2.45 moles/L / 32.04 g/mol)) results in approximately 2.70 m.

For the perchloric acid, the percent composition by mass can be calculated as % mass = (mol/L * g/mol * mL/g * 100) / (% by mass = (9.20 mol/L * 100.5 g/mol * 1 mL/1.54 g)*100).which gives you a result of approximately 60% by mass.

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