This is basically the lab I typed up tonight to hand in tomorrow. I'm wondering: 1. Does it look okay? 2. Is there any additional information I should add/anything to take out? 3. Is my third calculation totally off or is it just experimental error? Purpose: To calculate the molar enthalpy of the compound sodium hydroxide by using water in a calorimeter. Materials: An insulated coffee cup with a fitted lid Water (varying volumes in each trial) Solid sodium hydroxide (varying masses in each trial) A thermometer Procedure: 1. Fill the coffee cup with 100ml of water. Find and record the temperature fdsof the water. 2. Weigh 9.36g of sodium hydroxide. 3. Add the sodium hydroxide to the water and quickly place the lid on top of the cup. 4. Lift the lid a bit and stir with the thermometer until the sodium hydroxide dissolves. Record the temperature of the solution. 5. Pour out the solution and clean the coffee cup thoroughly. 6. Repeat steps 1-5 twice. In the second trial, use 100ml of water and 29.27g of sodium hydroxide. In the third trial, use 75ml of water and 71.52g of sodium hydroxide. Observations: Trial Volume of Water Mass of Sodium Hydroxide Temperature initial Temperature final 1 100ml 9.36g 22°C 40°C 2 100ml 29.27g 23°C 70°C 3 75ml 71.52g 22°C 85°C Calculations: ΔHNaOH=qwater nH=vcΔt H=vcΔt n For trial 1 H= 0.1L* 4.19kJ *18°C *40g L ● °C * 9.36g * 1 mol = 32.23 kJ/mol For trial 2 H= 0.1L* 4.19kJ *47°C* 40g L ● °C *29.27g *1 mol =26.91 kJ/mol For trial 3 H= 0.075L * 4.19kJ * 63°C * 40g L ● °C *71.52g *1 mol =11.07 kJ/mol Conclusions: I found that the third trial seemed to be abnormal and so I won’t include it in an average. By averaging the two closer enthalpies, I got 29.57 kJ/mol as the molar enthalpy for sodium hydroxide. Thanks in advance! EDIT: Sorry if the math and the charts are a little illegible. It's hard to communicate math over the internet. And I think I may have posted it in the wrong spot originally, sorry!