Question

23) 22.22 g of MgBr, are dissolved in in 1000 g of water solution. What is the new: a. Freezing point of this solution (2pts) b. Melting point of this solution (2pts)

Answer 1

PAGE DATE: 23) 22.229. of MgBo2 dissolved in 3000g water solution a) S To find freezing point of AT = Kemo solution, we ruse - where, AT = difference berueen Heup. Of Solution Kp = freezing point depression constant m = molality i = vaut noth fallor. li.e. no. of ions performing seaction). m (molality) = No. Of moles of solule. (mg382) Ell kg of solvent (ie. H₂O) no. of moles in 22.229 Mg B8 = 22229*_1 molar mass of Mg BOZ - 22-229 x mol ... 184-113g. - 0.1206 moles. Capprox.) m = = 0. 2 0.1206 1 l mass in kg : 1000g H0 - 1000g x kg To Jan 1000g - 1 kg. mol o kg: = weg AT = 11.86c kg (0.12 mol (3) - mol kg AT = 0.6696 °C ( change in temperature) so freezing point of solution will be - 10-0.6696) c = -0.6696 "Coon - 7 Freezing point of solution is - 0.669650 •

To find boiling point of solution, we use AT = Kumii where Kb = boiling point elevation constant (of water) m = molality E vant noff factor oud Sini TS m = no. of moles of solute cle Mg B82) kg of solvent Cie H₂O) moles of 22.22g mgB%, x mot 184-113 g = 0.1206 moles mass in kg 1000g H₂O x kg = 1kg 1000g Em 0.12 mol 0.1206 mol = low 1 kg kg AT=/0.51°C kg 1) (0-12 mol (3) mol = 0·1836°C • Boiling point = Old Boiling point (water) + AT = (100+ 0.1836) °C Boiling point = 100. 1836°C doo - TN INOTE: melting formula to calculate There point is no such ..