Question

una muestra de la sustancia con una masa de libera cal al congelarse en su punto de congelación. si la sustancia tiene una masa molar de, ¿cuál es el calor de fusión molar de la sustancia? utilice x326.0 g4325.8x58.45 g/molx q = nδh 775.6 cal/mol 13.31 cal/mol 19054.7 cal/mol 74.00 cal/mol

Explanation:

1. First, assume the heat released \(q = 4325.8\) cal, mass \(m=326.0\) g and molar - mass \(M = 58.45\) g/mol.

• Calculate the number of moles \(n\) of the substance using the formula \(n=\frac{m}{M}\).
• Substitute \(m = 326.0\) g and \(M = 58.45\) g/mol into the formula: \(n=\frac{326.0}{58.45}\text{mol}\approx5.58\) mol.

1. Then, use the formula \(q = n\Delta H\) to find the molar heat of fusion \(\Delta H\).

• Rearrange the formula \(\Delta H=\frac{q}{n}\).
• Substitute \(q = 4325.8\) cal and \(n\approx5.58\) mol into the formula: \(\Delta H=\frac{4325.8}{5.58}\text{cal/mol}\approx775.6\) cal/mol.

Answer:

775.6 cal/mol