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Topic: Calculating the Boiling Point of a Solution of NaCl in water  (Read 31744 times)

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afchick7689

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Calculating the Boiling Point of a Solution of NaCl in water
« on: August 29, 2005, 06:04:57 PM »
Can someone check my work here?

The question read:Calculate the boiling point of a solution of 10 grams of sodium chloride in 200 grams of water.

10 g NaCl + 200 g H2O (NaCl= solute; H2O = solvent)

200 g  H2O   |    1 kg H2O   = .2 kg H2O
      |  1000 g H2O

10 g NaCl    |     1 mol NaCl    = 0.1709 moles NaCl
     |    58.5 g NaCl

DT= i * Kb * m

i = for NaCl it equals 2

Kb = constant for the solvent; for water it equals 0.52 °C

m = molality (moles of solute/ kg of solvent)  
 
m=0.1709 moles NaCl/ .2 kg H2O= 0.8545 moles/kg= 0.8545m

DT= (2) (0.52 °C ) 0.8545 m = 0.88868 oC (W/ SIG FIGS DT = 0.9 oC)+100 oC= 100.09 oC) (Answer w/ SIG FIGS = 100 oC)

Offline Borek

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Re:Calculating the Boiling Point of a Solution of NaCl in water
« Reply #1 on: August 29, 2005, 06:27:00 PM »
Seems OK with me, but second opinion will not make any harm.

Anyone? Molality is calculated correctly (CASC rulez :) ).
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Offline sdekivit

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Re:Calculating the Boiling Point of a Solution of NaCl in water
« Reply #2 on: August 30, 2005, 11:16:49 AM »
Can someone check my work here?

The question read:Calculate the boiling point of a solution of 10 grams of sodium chloride in 200 grams of water.

10 g NaCl + 200 g H2O (NaCl= solute; H2O = solvent)

200 g  H2O   |    1 kg H2O   = .2 kg H2O
???   |  1000 g H2O

10 g NaCl    |     1 mol NaCl    = 0.1709 moles NaCl
???  |    58.5 g NaCl

DT= i * Kb * m

i = for NaCl it equals 2

Kb = constant for the solvent; for water it equals 0.52 °C

m = molality (moles of solute/ kg of solvent)  
 
m=0.1709 moles NaCl/ .2 kg H2O= 0.8545 moles/kg= 0.8545m

DT= (2) (0.52 °C ) 0.8545 m = 0.88868 oC (W/ SIG FIGS DT = 0.9 oC)+100 oC= 100.09 oC) (Answer w/ SIG FIGS = 100 oC)


seems ok :) only that 100 + 0,9 = 100,9 thus the answer is 101 degrees C.
« Last Edit: August 30, 2005, 11:18:37 AM by sdekivit »

Blueshawk

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Re:Calculating the Boiling Point of a Solution of NaCl in water
« Reply #3 on: August 30, 2005, 12:26:09 PM »
For me...when I look at the sig figs....the lowest decimal place holder is 2 places with 0.52.

So the final temp ... I would report as either 100.90 C   or 101.00 C.  + or - 0.10 C

But that is just me.  :)
« Last Edit: August 30, 2005, 12:27:21 PM by Blueshawk »

afchick7689

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Re:Calculating the Boiling Point of a Solution of NaCl in water
« Reply #4 on: August 30, 2005, 04:40:54 PM »
Ooo very true, thanks so much for the help... but i think the answer has to be in 1 sig fig (because the data given for the grams of NaCl had only 1 sig fig), so how would i write the answer 100.9 (or 101) with only 1 sig fig?? wouldnt this just round down to 100 again because of the sig figs?

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Re:Calculating the Boiling Point of a Solution of NaCl in water
« Reply #5 on: August 30, 2005, 06:30:08 PM »
Ooo very true, thanks so much for the help... but i think the answer has to be in 1 sig fig (because the data given for the grams of NaCl had only 1 sig fig), so how would i write the answer 100.9 (or 101) with only 1 sig fig?? wouldnt this just round down to 100 again because of the sig figs?

Good point, but I will write 100.9 without hesitation.

100.0 is an exact number. Result of your calculation is 0.9 - and it is delta T, temperature elevation to be expected. Thus writing the result down as 100.9 you are not adding precision to the final result.

The problem is, significant digits are far from being precise. Your result - to be correct - should look like 100.9 +/- error value.
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