Question

5. Molality, m =- The molality of NaBr solution is 1.25 m 1.25 mole of NaBr I 1.0 kg of the solvent Question 17. Calculate the number of g solvent that contains 12 g NaBr of 1.25 m solution of NaBr. [8.14E2 g solvent] Question 18. Calculate the molality of a solution containing 17.2 g of ethylene glycol (CzH02 ) dissolved in 0.500 kg water. [5.54E-1 m) Question 19. What is the formula of sucrose? Calculate the molality of a sucrose solution containing 50.4 g sucrose and 0.332 kg water. [ 4.43 E-1 m)

Answer 1

molality = moles of solute kg of solvent mars meles mass molar mass 17) 1.25m = m of NaBr molar mass of Na Br x kg of solvent 1.25m = 12 g 102.894 g/mol x Msolvent insolvent = hores quegel 12 g - 102.894 g/mid x 1.25m m solvent = 0.0933 kg 18) Note: 8.148 10 ² g is wrong. molality = 17:29 62.07g/ml x ois kg = 55.4m = 5.54x10 m. 19) formula of sucrose C12 M22 Oll Molar mass = 342.2965 g/md molality = 50.49 34 2 -2965g/mol X0.332 kg = 0.443m - 4.43xlo'm