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What is the boiling point of a solution that contains 3 moles of KBr in 2000g of...

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connorcolin3 | Valedictorian

Posted May 25, 2013 at 9:09 PM via web

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What is the boiling point of a solution that contains 3 moles of KBr in 2000g of water?

a) 97 Degrees C

b) 99.7 Degrees C

c)101.4 Degrees C

d) 103 degrees c

 

 

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llltkl | College Teacher | Valedictorian

Posted May 26, 2013 at 12:23 AM (Answer #1)

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When a nonvolatile esolute is dissolved in water, according to Raoult’s laws, there will be an elevatio in the boiling point of water. The extent of elevation is given by

 `DeltaTb = i*Kb *m` , where i is the number of species obtained from the dissociation (or association)of one molecule of the solute in solution, Kb is a constant, called molal ebullioscopic constant for a particular solvent, and m is the molal strength of the solute.

For water, Kb = 0.512,

Molality is the number of moles of solute dissolved per 1Kg solvent.

Here, 3 moles of KBr is dissolved in 2Kg water, therefore molality = 3/2 = 1.5.

KBr is an electrolyte, completely dissociating in aqueous solution as

KBr (s)→ K+(aq.) + Br-(aq.)

Hence i = 2.

Putting these values in equation,

ΔTb = 2*0.512 *1.5

= 1.536

Therefore, boiling point of the solution would be 100+1.536 = 101.5 degrees (approximately).

Correct answer is therefore, option c)101.4 Degrees C.

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