Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: zerality on September 22, 2006, 03:00:15 PM
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Hi,
I am getting confused as how to calculate molarity in a lab experiment performed in class.
We had to determine the concentration of solids, reported as NaCl, in a sample of seawater by:
determine density and mass of a sample of seawater. we boiled the seawater to dryness and measured the mass of the residual solids. from the data we had to calculate the % by mass, M, and m of the seawater sample.
i managed to calculate % by mass (3.12%) and the molality, but am confused as how to calculate M.
to calculate m i assumed a 1000g solution which would then be composed of 31.2 grams of NaCl and 1000-31.2 grams (968.8 g) H2O. Therefore molality would equal 31g/.9688 kg = 32 g NaCl/Kg H20 = 32/58.44 = .55 m
To calculate molarity: I said density = mass/voulme; mass (solution) = 1.020g/ml x 1000ml = 1020 g
since % by mass of NaCl is 3.12%, in 1000ml the mass = (1020 x 3.12)/(100) = 31.82g NaCl.
therefore the moles of NaCl in 1 liter of solution is 31.82/58.44 = .54 M= .54
Is my answer for M correct?
Many Thanks
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you know the density of the solution: 1,02 g / mL and 3,12 mass% = NaCl
Now it's easier to take 100 g seawater, since percetage means: per hundred.
know 100g seawater thus contains 3,12 g NaCl and using the density, you can calculate the volume of 100g seawater.
Then you need to reconsider the definition of molarity: the amount of mol solute per L solvent.
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Thanks for the response.
Isn' the definition of molarity: moles solute/liter of solution. If so, isn't my answer correct?
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Molarity is defined as moles of solute per liter of solution
http://en.wikipedia.org/wiki/Molarity#Molarity
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http://www.chembuddy.com/?left=concentration&right=molarity
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So is my answer correct?
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It's right if you don't need to worry about significant figures.