Question

Enter your answer in the provided box. The formula that governs the depression of freezing point and elevation of boiling point for a solution consisting of a solute dissolved in a solvent is: AT=ix ky Xm Book where: AT = the temperature change between a pure solvent and its solution erences i = the number of species per mole of solute that are dissolved in the solvent (e.g., i=1 for a non-ionic solute that does not break apart into ions; i = 2 for an ionic solute such as KBr that breaks apart into two ions, K and Br", and so on) k = a constant related to the solvent m=molality of the solute (molality, m = moles of solute /kg of solvent) Using the above equation, how many grams of salt (CaCl) would need to be added to 1 L of water in order for the boiling point of the solution to reach 109.5 °C? Assume that the density of water is 1.0 g/mL and that CaCl, completely dissociates into three ions - i.e., a Cation and two C ions. The boiling point constant, kb, for water is 0.515°Cm. Report your answer to the nearest 0.1 g.

solve please.

Answer 1

m Molarity of the solute Kb 0.515 °C/m ATb = Change in boiling point ATb 109.5 100 = 9.5 °C AТЬ 3 і К, m дТЬ m = i Kb 9.5 m = 3 X 0.515 - 6.15 M moles of solute Molality Kg of solvent Moles of solute 6.14 Moles of solute = 6.14 Molar mass of CaC12 110.98 g/mol Mass of CaC12 Moles x molar mass = 6.14 x 110.98 -681.41 g