Reference no: EM13151165

Compute the mass of solute and water needed to prepare the following solutions: 1. 150 g of 1.00 m (NH4)2SO4: (a) from pure solid (NH4)2SO4 (b) from solid (NH4)2SO4 with a purity of 87.2 wt-% 2. 1.50 L of a 20 wt-% Pb(NO3)2 solution whose density is 1.20 g/mL (a) from pure solid Pb(NO3)2 (b) from solid Pb(NO3)2 with a purity of 98.9 wt-% This is the work I've done so far (please check): 1. a) Molality = (moles of solute) / (kilograms of solvent) moles of (NH4)2SO4 = 150 g (NH4)2SO4 × (1 kg)/(1000 g) × (1.00 mol (NH4)2SO4) / (1 kg (NH4)2SO4) =0.15 mol (NH4)2SO4 mass of (NH4)2SO4 = 0.15 mol (NH4)2SO4 × (132.14 g (NH4)2SO4) / (1 mol (NH4)2SO4) = 19.8 g (NH4)2SO4 mass of solution = mass of solute + mass of solvent mass of H2O = 150 g soln - 19.8 g (NH4)2SO4 = 130.2 g H2O (b) Mass % of component = (mass of component in soln) / (total mass of soln) × 100 mass of (NH4)2SO4 = (19.8 g (NH4)2SO4) / 0.872 = 22.7 g impure (NH4)2SO4 2. a) total mass of Pb(NO3)2 = (1.20 g Pb(NO3)2) / (mL Pb(NO3)2) × (1000 mL)/(1 L) × 1.50 L soln = 1800 g Pb(NO3)2 Mass % of component = (mass of component in soln) / (total mass of soln) × 100 mass of Pb(NO3)2 = 0.20 × 1800 g Pb(NO3)2 = 360.g Pb(NO3)2 How do I determine the mass of water for this question? (b) Mass % of component = (mass of component in soln) / (total mass of soln) × 100 mass of Pb(NO3)2 = (360.g Pb(NO3)2) / 0.989 = 364.0 g impure Pb(NO3)2 How do I determine the mass of water for this question?