In Part 2 of Lab #12 you will dissolve about 3 g of urea in 50.0 mL of water and
ID: 989941 • Letter: I
Question
In Part 2 of Lab #12 you will dissolve about 3 g of urea in 50.0 mL of water and use the change in temperature to determine H°soln. Let’s go through the calculations. Suppose you dissolve 3.77 g of a solid in 50.0 mL of water. You note that the temperature changes from 25.0 °C to 28.1 °C. Part 1: What is the mass of the solution? Assume the density of water is 1.000 g/mL. Part 2: Determine the absolute value of the change in temperature of the solution. Part 3: Determine the total heat involved in dissolving your samples of the solid in 50.0 mL of water. Report your answer with the proper sign. Part 4: If the solid has a molar mass of 85.9 g/mol, determine H°soln for the solid. Report your answer with the proper sign.
Explanation / Answer
In the colorimetric methods of enthalpy measurement
Q=m *Cp *deltaT
1.M is mass of the water in the calorimeter in grams , or the mass of whatever substance is acting as the environment. In reality we should assume that the material part of the calorimeter also absorbs heat. But in this course, we usually ignore that part.So here we consider only the environment hence mass is 50 g (since density of water is 1 g/mol)
2.Absolute change in the temparature is delta T = T2 - T1 = 3.1 C
3. Total heat Q=50g * 4.179j/g C* 3.1 C ( 4.179 is specific heat of water)
Q = 647.745 J
4. molar enthalpy delta H = -Q/n
n=3.77g/85.9 g/mol (mass/molar mass=moles)
= 0.04388
therefor delta H = -647.745/0.04388 J/mol
=-14761.73 J = -14.761 KJ/mol
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