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Title: What is the molality of particles in a solution containing 13.9 g of KBr dissolved in 760. g of water? Post by: dave12 on Mar 20, 2013 When i tried this problem i got .154 which is wrong. Could it be the difference between molarity of a solution and molarity of particles in the solution?
Title: What is the molality of particles in a solution containing 13.9 g of KBr dissolved in 760. g of water? Post by: smitty2 on Mar 20, 2013 MW = 119.02 KBr
molality = (13.9/119.02)/0.760 = 0.153 molal Title: What is the molality of particles in a solution containing 13.9 g of KBr dissolved in 760. g of water? Post by: Smitty on Mar 20, 2013 Molality is needed here because by separating out the solute from the solvent we can eliminate the temperature dependency within the volume portion of molarity. This is useful for the colligative properties of freezing point depression and boiling point elevation.
Molality is moles solute per kilogram of solvent. (13.9 g / 119.01 g/mol) / 0.760 kg = 0.154 mol/kg -or- 0.154 molal. Now this is molality of KBr. Since KBr is an ionic substance, it will dissociate into its component ions in aqueous solution. If we assume complete dissociation, then each mole of KBr produces two moles of particles, a mole of K+ ions and a mole of Br- ions. So, the molality of particles is 0.154 x 2 = 0.308 molal. However, it is doubtful that all of the KBr in a solution of this concentration would be completely dissociated, so the effective concentration, or activity of the particles would be something less than 0.308 molal, possibly as low as 0.25 molal would be observed. |