Question

Given the information shown in process "B", calculate the molar enthalphy of fusion for ice

A) 40.1 kJ/mol
B) 4.50 kJ/mol
C) 18.0 kj/mol
D) 6.02 kJ/mol

Answer 1

D) 6.02 kJ/mol

Step-by-step explanation:

Hello,

In this case, since 200kJ melted 1.5 kg of ice (2.0kg-0.5kg), we can compute the melted moles:

`n=(1.5kg)/(18kg/kmol) *(1000mol)/(1kmol)=83.33mol`

Then, we compute the molar enthalpy of fusion by diving the melted moles to the applied heat:

`\Delta H_(fusion)=(500kJ)/(83.33mol)\\ \\\Delta H_(fusion)=6.02kJ/mol`

Hence answer is D) 6.02 kJ/mol.

Best regards.