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Lowering the Freezing Point of Water
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tayjmc
- Posts: 1
- Joined: Tue Jun 02, 2015 1:59 pm
- Occupation: student tenth grade
- Project Question: I was looking at your website and was interested in completing your Lowering the Freezing Point of Water experiment. I was simply wondering if there is a certain reason as to why 11.7g of NaCl for example has been selected, instead of merely 12g. What is the reasoning behind the masses provided?
Many Thanks,
Taylor - Project Due Date: 24/06/15
- Project Status: I am finished with my experiment and analyzing the data
Lowering the Freezing Point of Water
I was looking at your website and was interested in completing your Lowering the Freezing Point of Water experiment. I was simply wondering if there is a certain reason as to why 11.7g of NaCl for example has been selected, instead of merely 12g. What is the reasoning behind the masses provided?
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deleted-140482
- Former Expert
- Posts: 186
- Joined: Fri Aug 09, 2013 12:56 pm
- Occupation: Postdoctoral Fellow
- Project Question: Signing up to be an Expert
- Project Due Date: n/a
- Project Status: Not applicable
Re: Lowering the Freezing Point of Water
Hi tayjmc,
Those numbers are chosen to make the molality of the solution even. Molality is defined as the number of moles of a substance, divided by the weight (in kg) of the solvent. You can read more about this in the background section of this project idea. To calculate the molality of a solution then, you will need to convert your grams of solute (salt in this case) into moles. This can be done by dividing the grams by the molecular weight of the substance (in the case of NaCl, or table salt, the molecular weight is 58.44g/mol). Then you would divide this number by the kg of solvent. For water, 1L = 1kg, so the 100mL, or 0.1L we are using here is equivalent to 0.1kg. Thus, to calculate the molality of a solution of 11.7g salt in 100mL water, you would first divide 11.7g by 58.44, to get 0.2 moles of NaCl. Then divide 0.2 moles NaCl by 0.1kg water to find that this makes a solution with a molality of 2. That is why 11.7g was chosen, rather than 12g. You can read a lot more about this in the background and procedure sections of this project idea. Take a look at this information, and think about why we would want to use the molality of a solution, rather than simply grams per mL.
I hope this helps!
JMP
Those numbers are chosen to make the molality of the solution even. Molality is defined as the number of moles of a substance, divided by the weight (in kg) of the solvent. You can read more about this in the background section of this project idea. To calculate the molality of a solution then, you will need to convert your grams of solute (salt in this case) into moles. This can be done by dividing the grams by the molecular weight of the substance (in the case of NaCl, or table salt, the molecular weight is 58.44g/mol). Then you would divide this number by the kg of solvent. For water, 1L = 1kg, so the 100mL, or 0.1L we are using here is equivalent to 0.1kg. Thus, to calculate the molality of a solution of 11.7g salt in 100mL water, you would first divide 11.7g by 58.44, to get 0.2 moles of NaCl. Then divide 0.2 moles NaCl by 0.1kg water to find that this makes a solution with a molality of 2. That is why 11.7g was chosen, rather than 12g. You can read a lot more about this in the background and procedure sections of this project idea. Take a look at this information, and think about why we would want to use the molality of a solution, rather than simply grams per mL.
I hope this helps!
JMP