Hey,
for chemistry class we are given days in the lab to conduct an experiment, then answer a number of questions that correlate with that experiment and how we can make use of it. This week we looked at the Molar Fusion of Ice and there are some questions i don't quite understand.
The experiment went like this: water was poured into a beaker, heated (temperature and volume was taken), then cooled with ice cubes (temperature/volume taken again).
Here is my table of information:
volume of water after heated:100ml
temperature after heated:55ºC
temperature after cooled:.5ºC
Volume after cooled:175ml
Here are the questions:
1) Determine the volume of water in the ice-water mix that came from the melted ice.
2)Determine the mass of the water that came from the melted ice. Assume the density of water is 1.00g/ml
3)Determine the change in temperature of the warm water
4) Determine the mass of the warm water
5) Calculate the energy released by the warm water as it cooled through ΔT degrees (Hint: specific heat of water is 4.184J/gºC)
6) Calculate the energy released by the warm water for every gram of ice melted
7) Calculate the molar heat of fusion of ice (the kilojoules required to melt one mole of ice)
8) The accepted value for the molar heat of fusion of ice is 6.02. What is the percent error in the experimental value?
My Answers:
1 through 5 are mostly self-explanatory, the rest I'm pretty sure I'm doing them wrong
1) 75ml
2) 75g
3) -54.5ºC
4) 100g
5) q=CmΔT
q = 4.184(100)(-54.5)
q= -22802.8J
6) -22802.8/75=-304J
7) -22.8028kJ /18g=1.26
8) (1.26-6.02)/6.02*100=79%
umm.. yah.. can anyone help?
Based on your description:
3) 55º-5º = 50º
5) (4.18J/g.ºC)(100.0g)(50º) = 20900 J
6) 20900J/75g = 278.7J/g
7) (278.7J/g)(18.0g/mol) = 5016 J/mole = 5.016 kJ/mol
8) %Error = (6.02-5.02)(100)/6.02 = 17%
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NOTE: Some of the heat lost by the warm water went into heating the melted ice from 0 deg C to 5.0 deg C. That was not taken into account in your procedure.
Well I wrote the value .5 for the lowest temperatures but I think I will be able to use the equations you used anyway, thanks for the help!
Sure! I'd be happy to help you with the remaining questions.
6) To calculate the energy released by the warm water for every gram of ice melted, you need to divide the total energy released (which you calculated in question 5) by the mass of ice melted.
You've correctly calculated the total energy released as -22,802.8 J. Now, you need to divide this by the mass of ice melted. The mass of ice melted is equal to the mass of water that came from the melted ice, which you found to be 75 g in question 2.
So, the calculation would be: -22,802.8 J / 75 g = -304.04 J/g (rounded to two decimal places)
Therefore, for every gram of ice melted, the warm water releases approximately -304 J of energy.
7) To calculate the molar heat of fusion of ice, you need to convert the energy released per gram of ice (which you calculated in question 6) into kilojoules per mole.
Since the molar mass of water is approximately 18 g/mol (1 mole of water contains 18 grams), you can use this conversion factor.
To convert from joules to kilojoules, divide the answer from question 6 (-304 J/g) by 1000.
So, the calculation would be: -304 J/g / 1000 = -0.304 kJ/g
Therefore, the molar heat of fusion of ice is approximately -0.304 kilojoules per gram.
8) To calculate the percent error in the experimental value compared to the accepted value, use the following formula:
Percent Error = [(Experimental Value - Accepted Value) / Accepted Value] * 100
Substituting the values, the calculation would be:
Percent Error = [(1.26 - 6.02) / 6.02] * 100 = -79.07% (rounding to two decimal places)
Therefore, the percent error in the experimental value is approximately -79.07%.
It seems like you may have made a sign error in your calculation for the percent error. The negative sign indicates that the experimental value is lower than the accepted value.
I hope this clears up your confusion and helps you with the remaining questions! Let me know if you have any further questions.