Question

the molar heat of fusion of platinum (pt) is 4.700 kcal/mol. how much heat must be added to 85.5 g of solid platinum at its melting point to completely melt it? use the formula ( q = n cdot delta h ). (\boxed{quad}) kcal

Explanation:

Step1: Find moles of Pt

Molar mass of Pt is 195.08 g/mol. Moles \( n = \frac{\text{mass}}{\text{molar mass}} = \frac{85.5\ \text{g}}{195.08\ \text{g/mol}} \approx 0.4383\ \text{mol} \)

Step2: Calculate heat using \( q = n \cdot \Delta H \)

Given \( \Delta H = 4.790\ \text{kJ/mol} \) (assuming kJ, likely typo for kJ? Or kcal? Wait, molar heat of fusion for Pt is ~22.1 kJ/mol, but maybe problem has 4.790 kJ/mol. Wait, let's proceed with given. \( q = 0.4383\ \text{mol} \times 4.790\ \text{kJ/mol} \approx 2.10\ \text{kJ} \)? Wait, maybe unit is kcal? Wait, no, Pt's molar heat of fusion is ~22 kJ/mol. But following problem's numbers: \( n = 85.5 / 195.08 ≈ 0.438 \), \( q = 0.438 \times 4.790 ≈ 2.10 \) (if kJ) or if kcal, but maybe problem has 4.790 kJ/mol. Wait, maybe I misread: "molar heat of fusion of platinum (Pt) is 4.790 kJ/mol". Then:

\( n = \frac{85.5\ \text{g}}{195.08\ \text{g/mol}} \approx 0.4383\ \text{mol} \)

\( q = 0.4383\ \text{mol} \times 4.790\ \text{kJ/mol} \approx 2.10\ \text{kJ} \)? Wait, no, 0.4383 4.790 ≈ 2.10? Wait 0.44.79=1.916, 0.0383*4.79≈0.183, total≈2.099≈2.10 kJ. But maybe the problem's molar heat is 4.790 kcal/mol? No, Pt's is ~22 kJ/mol. But regardless, following the problem's numbers:

Wait, 85.5 g Pt. Molar mass 195.08 g/mol. Moles: 85.5 / 195.08 ≈ 0.438 mol. Then q = 0.438 mol * 4.790 kJ/mol ≈ 2.10 kJ. But maybe the answer is around 2.10 kJ (or kcal? But units in problem: "kJ" or "kcal"? The box has "kJ" or "kcal"? The problem says "kJ" maybe. Wait, let's recalculate:

85.5 / 195.08 = 0.4383 mol.

0.4383 4.790 = 0.43834 + 0.4383*0.790 = 1.7532 + 0.3462 = 2.0994 ≈ 2.10 kJ.

Answer:

\boxed{2.10} (assuming kJ, or adjust if unit is different, but based on calculation)