Question

Zn(s) + 2H +(aq) --> Zn 2+(aq) + H2(g). . A sample of zinc is placed in the ice calorimeter described in the \"experimental\" section below. If 0.0657g of zinc causes a decrease of 0.109mL in the ice/water volume of the calorimeter, what is the enthalpy change , per mole of Zinc, for the above reaction per mole of zinc.. . It is also given that enthalpy change for process H20(s) -> H20(l) is +6.01kJ/mol or 333J/g and it can be determined that 3.68 kJ are released per mL change in volume of ice/water mixture..

Answer 1

H2O(s) ==> H2O(l) 
The density of liquid H2O = 1.00 g/mL. 
The density of ice is 0.917 g/mL. 
Suppose we start with 1 mol (18 g) H2O(s) and it melts to the liquid. 
As a liquid it will occupy 18g/1.00 = 18 mL. 
As a solid it will occupy 18/0.917 = 19.63 mL. 
19.63-18.0 = 1.63 mL diffeence. 
The heat fusion of ice is 6.01 kJ/mol; therefore, 6.01/1.63 = about 3.69 kJ for each mL difference. 

In this experiment we have a difference of 0.109 mL; then 3.69 kJ/mL x 0.109 mL = 0.402 kJ for 0.0657 g Zn. You want per mol Zn and you have 0.0657/65.39 = about 0.001 mol Zn so 
0.402 kJ/0.001 = 402 kJ/mol Zn