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# What is the molality of particles in a solution containing 13.9 g of KBr dissolved in 760. g of water?

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10 years ago
 What is the molality of particles in a solution containing 13.9 g of KBr dissolved in 760. g of water? 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? Read 371 times 2 Replies

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wrote...
10 years ago
 MW = 119.02 KBrmolality = (13.9/119.02)/0.760            =  0.153 molal
wrote...
10 years ago
 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.
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