Ice cream making: lowering the freezing point

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ilovescience37
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Joined: Tue Sep 29, 2015 5:15 pm
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Ice cream making: lowering the freezing point

Post by ilovescience37 »

why does it need to be that exact grams of the salt and sugar??

Test liquid #2 = 5.8 g NaCl in 100 mL water
Test liquid #3 = 11.7 g NaCl in 100 mL water
Test liquid #4 = 17.1 g sucrose in 100 mL water
Test liquid #5 = 34.2 g sucrose in 100 mL water
Test liquid #6 = 68.5 g sucrose in 100 mL water

please is urgent
norman40
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Re: Ice cream making: lowering the freezing point

Post by norman40 »

Hello ilovescience37,

As explained in the Background section of the Chemistry of Ice Cream Making project, the equation describing freezing point depression is in terms of molality. The Background and Procedure sections explain how to calculate concentrations in molality.

The weights of NaCl and sugar listed in the Procedure section were chosen to give the same molality range for both compounds. For example, when you finish your calculations you’ll see that 2.9 g NaCl in 100 mL of water is the same molality as 17.1 g sucrose in 100 mL of water.

I hope this helps. Please post again if you have more questions.

A. Norman
ilovescience37
Posts: 9
Joined: Tue Sep 29, 2015 5:15 pm
Occupation: Student

Re: Ice cream making: lowering the freezing point

Post by ilovescience37 »

ok ,but first

is there any specific formula to calculate the moles and if so can someone please explain me the whole 3 equations things please!!!!
norman40
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Joined: Mon Jul 14, 2014 1:49 pm
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Re: Ice cream making: lowering the freezing point

Post by norman40 »

Hello ilovescience37,

The number of moles is the weight of a substance divided by its gram molecular weight. For example, say you have 2.9 g NaCl. The gram molecular weight of NaCl is 58.4 g/mole. So the number of moles is 2.9/58.4 or 0.05 moles of NaCl. There’s more information about this in the Procedure section under the heading “Calculation of Expected Freezing Point Depression”.

Equation 1 is about calculating molality. Divide the number of moles you have by the weight of solvent. In this experiment you have 100 mL of water as the solvent. The molality calculation uses solvent weight (in kg) so you have to convert the 100 mL to weight. The density of water is 1 g/mL so 100 mL weighs 0.1 kg.

Equation 2 is about how the freezing point of a liquid changes with molality. The equation predicts that the freezing point depression will increase when the molality of the solution increases. There is a factor in the equation (van’t Hoff factor) that differs according to the solute (NaCl or sugar). Solutes with a larger van’t Hoff factor will give a greater freezing point depression.

Equation 3 shows how to calculate freezing point from freezing point depression.

I hope this helps. Please post again if you have more questions.

A. Norman
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