0 votes
85 views
in ThermoChemistry by (430 points)
A student wants to measure the heat of crystallization of sodium acetate trihydrate. The student takes 50 g of sodium acetate trihydrate and dissolves it in 100 ml of water at 70°C. The solution is then allowed to cool to 20°C, during which the sodium acetate trihydrate crystallizes out. The student measures the temperature change during this process and finds that the temperature drops

1 Answer

0 votes
by (410 points)
To calculate the heat of crystallization of sodium acetate trihydrate, we need to first determine the amount of heat released during the crystallization process. We can do this using the formula:q = mcTwhere q is the heat released, m is the mass of the solution, c is the specific heat capacity of the solution, and T is the change in temperature.The mass of the solution  m  can be calculated by adding the mass of sodium acetate trihydrate  50 g  to the mass of water  100 g . The specific heat capacity of water  c  is 4.18 J/gC.m = 50 g + 100 g = 150 gT = 45C - 20C = 25CNow we can calculate the heat released  q :q =  150 g  4.18 J/gC  25C  = 15675 JNext, we need to determine the number of moles of sodium acetate trihydrate in the 50 g sample. The molar mass of sodium acetate trihydrate  NaC2H3O23H2O  is:Na: 22.99 g/molC: 12.01 g/molH: 1.01 g/molO: 16.00 g/molMolar mass =  1  22.99  +  2  12.01  +  3  1.01  +  2  16.00  +  3  2  1.01  +  3  16.00  = 82.03 + 36.03 + 18.00 = 136.06 g/molNow we can calculate the number of moles of sodium acetate trihydrate:moles = mass / molar mass = 50 g / 136.06 g/mol = 0.367 molFinally, we can calculate the heat of crystallization per mole of sodium acetate trihydrate:Heat of crystallization = q / moles = 15675 J / 0.367 mol = 42710 J/molSince we want the answer in kJ/mol, we need to convert J/mol to kJ/mol:Heat of crystallization = 42710 J/mol   1 kJ / 1000 J  = 42.71 kJ/molSo, the heat of crystallization of sodium acetate trihydrate is approximately 42.71 kJ/mol.

Related questions

Welcome to Sarvan Science Q&A, where you can ask questions and receive answers from other members of the community.
...