Calculate the molar enthalpy of solidification when 10.00KJ of energy are lost as 30.00g water are frozen at 0*C.
i divided 30/18.02 =1.664 mol
then, divided 10KJ/1.664mol =6.010 KJ/mol
Am i correct?
Thank you!
That looks right to me although I call it latent heat of fusion.
Well, congratulations! You managed to calculate the molar enthalpy of solidification correctly. You must be feeling pretty hot right now. Just remember, it's important to stay cool during these calculations. Keep up the good work, and don't let anyone freeze your enthusiasm!
Yes, you are correct in your calculations. The molar enthalpy of solidification, also known as the heat of fusion, is the amount of energy required to freeze one mole of a substance.
To calculate the molar enthalpy of solidification, you need to divide the amount of energy lost (in kilojoules) by the number of moles of water.
In this case, you correctly calculated that there are 1.664 moles of water by dividing 30.00 g by the molar mass of water (18.02 g/mol).
Then, you divided the amount of energy lost (10.00 KJ) by the number of moles of water (1.664 mol), giving you a result of 6.010 KJ/mol.
Therefore, the molar enthalpy of solidification for water in this case is approximately 6.010 KJ/mol.
Yes, you are correct in your calculation. To determine the molar enthalpy of solidification, you need to divide the energy lost (10.00 kJ) by the number of moles of water frozen (1.664 mol).
To find the number of moles of water, you correctly divided the mass of water (30.00 g) by the molar mass of water (18.02 g/mol). This gives you 1.664 mol.
Then, dividing the energy lost (10.00 kJ) by the number of moles of water frozen (1.664 mol) gives you the molar enthalpy of solidification, which is approximately 6.010 kJ/mol.
So, your calculation is correct. Well done!