STUDY QUE STION Consider naphthalene, C_(10)H_(8) the active ingredient of moth balls. A solution of naphthalene is prepared by mixing 25.0 g of naphthalene with 0750 L of carbon disulfide, CS_(2)(d=1.263g/mL ). Assume that the volume of the solu- tion is the same as that of the solvent after the solution is prepared. (a) What is the mass percent of naphthalene in the solution? (b) What is the concentration of naphthalene in parts per million? (c) What is the density of the solution? (d) What is the molarity of the solution? Answers (e) What is the molality of the solution? (a)2.57% (b) 2.57times 10^4ppm (c) 1.30g/mL

Let's go through each part of the question step by step:<br /><br />(a) Mass percent of naphthalene in the solution:<br />Mass percent = (mass of solute / total mass of solution) * 100<br />Total mass of solution = mass of solute + mass of solvent<br />Total mass of solution = 25.0 g + (750 mL * 1.263 g/mL) = 25.0 g + 946.5 g = 971.5 g<br />Mass percent = (25.0 g / 971.5 g) * 100 = 2.57%<br /><br />(b) Concentration of naphthalene in parts per million (ppm):<br />ppm = (mass of solute / mass of solvent) * 10^6<br />ppm = (25.0 g / (750 mL * 1.263 g/mL)) * 10^6 = 2.57 * 10^4 ppm<br /><br />(c) Density of the solution:<br />Density = mass / volume<br />Density = total mass of solution / volume of solution<br />Density = 971.5 g / 750 mL = 1.30 g/mL<br /><br />(d) Molarity of the solution:<br />Molarity = moles of solute / volume of solution in liters<br />Moles of naphthalene = 25.0 g / (128.17 g/mol) = 0.1956 mol<br />Molarity = 0.1956 mol / 0.750 L = 0.2618 M<br /><br />(e) Molality of the solution:<br />Molality = moles of solute / mass of solvent in kilograms<br />Molality = 0.1956 mol / (0.750 kg) = 0.2618 m<br /><br />Therefore, the answers are:<br />(a) 2.57%<br />(b) 2.57 * 10^4 ppm<br />(c) 1.30 g/mL<br />(d) 0.2618 M<br />(e) 0.2618 m