Question
2. (a) The osmotic pressure of a solution containing 4.00gdm^-3 of PVC in dioxane is 65Nm^-2 at 20^circ C Calculate the number of PVC monomers (b) A solution containing 28.145g of R in 250g of water froze at -3.490^circ C determine the molecular mass of R
Answer
3.4
(304 Votes)
Seymour
Master · Tutor for 5 years
Answer
(a) 1.62 x 10^21 PVC monomers, (b) 58.64 g/mol
Explanation
(a) The osmotic pressure formula is π = n/VRT, where π is the osmotic pressure, n is the number of moles, V is the volume, R is the gas constant, and T is the temperature in Kelvin. We know that the osmotic pressure π is 65 Nm^-2, the concentration of PVC is 4.00 g/dm^3, the temperature T is 20°C (or 293.15 K), and the gas constant R is 8.314 J/(mol*K). We can rearrange the formula to solve for n, the number of moles: n = πV/RT. Since the volume V is not given, we can assume it to be 1 dm^3 (or 1 L) for simplicity. Substituting the given values into the formula, we get n = (65 * 1) / (8.314 * 293.15) = 0.0027 moles. Since the molar mass of PVC is 125 g/mol, the number of PVC monomers is 0.0027 * 6.022 x 10^23 = 1.62 x 10^21.(b) The freezing point depression formula is ΔTf = Kf*m, where ΔTf is the freezing point depression, Kf is the cryoscopic constant, and m is the molality of the solution. We know that the freezing point depression ΔTf is 3.49°C, the cryoscopic constant Kf for water is 1.86 °C/m, and the mass of the solvent (water) is 250g (or 0.25kg). We can rearrange the formula to solve for m, the molality: m = ΔTf / Kf = 3.49 / 1.86 = 1.92 mol/kg. The number of moles of R is therefore 1.92 * 0.25 = 0.48 moles. Since the mass of R is 28.145g, the molar mass of R is 28.145 / 0.48 = 58.64 g/mol.