The heat of fusion of pure silicon is $43.4 \mathrm{~kJ} / \mathrm{mol}$. How much energy would be needed to melt a $4.34-\mathrm{g}$ sample of silicon at its melting point of $1693 \mathrm{~K}$ ? Energy $=$ kJ
Answer
Rounding to two decimal places, the energy required to melt a \(4.34-\text{g}\) sample of silicon at its melting point of \(1693 \, \text{K}\) is approximately \(\boxed{6.71 \, \text{kJ}}\).
Step 1 ：Given that the heat of fusion of pure silicon is \(43.4 \, \text{kJ/mol}\), we need to find out how much energy would be needed to melt a \(4.34-\text{g}\) sample of silicon at its melting point of \(1693 \, \text{K}\).
Step 2 ：First, we need to convert the mass of silicon from grams to moles. We know that the molar mass of silicon is approximately \(28.0855 \, \text{g/mol}\).
Step 3 ：By dividing the mass of the silicon sample by the molar mass of silicon, we get the number of moles of silicon. So, \(\frac{4.34}{28.0855} = 0.15452813729504547 \, \text{mol}\).
Step 4 ：Next, we multiply the number of moles by the heat of fusion to find the total energy required to melt the sample. So, \(0.15452813729504547 \times 43.4 = 6.706521158604973 \, \text{kJ}\).
Step 5 ：Rounding to two decimal places, the energy required to melt a \(4.34-\text{g}\) sample of silicon at its melting point of \(1693 \, \text{K}\) is approximately \(\boxed{6.71 \, \text{kJ}}\).
