/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 101 It requires \(17.10 \mathrm{kJ}\... [FREE SOLUTION] | 91Ó°ÊÓ

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It requires \(17.10 \mathrm{kJ}\) to melt \(1.00 \times 10^{2} \mathrm{g}\) of urethane $\left[\mathrm{CO}_{2}\left(\mathrm{NH}_{2}\right) \mathrm{C}_{2} \mathrm{H}_{5}\right]\( at \)48.7^{\circ} \mathrm{C} .$ What is the latent heat of fusion of urethane in \(\mathrm{kJ} / \mathrm{mol} ?\)

Short Answer

Expert verified
Answer: The latent heat of fusion of urethane is 15.05 kJ/mol.

Step by step solution

01

Calculate the molar mass of urethane

To find the molar mass of urethane, \(\mathrm{CO}_{2}\left(\mathrm{NH}_{2}\right) \mathrm{C}_{2} \mathrm{H}_{5}\), we will sum the molar masses of each element in the compound. Molar mass of urethane, \(M_{urethane} = M_C + 2M_O + 2M_N + 7M_H\) Plugging in the molar masses of C, O, N, and H, we have \(M_{urethane} = 12 + 2(16) + 2(14) + 7(1) = 88 \mathrm{g/mol}\).
02

Calculate the moles of urethane

We need to find the number of moles given the mass of urethane, which is \(1.00 \times 10^{2} \mathrm{g}\). We will use the molar mass we calculated in step 1 to convert grams to moles. Moles of urethane, \(n = \frac{mass}{M_{urethane}} = \frac{1.00 \times 10^{2}\mathrm{g}}{88 \mathrm{g/mol}}\) \(n = 1.136 \mathrm{mol}\) (rounded to 3 decimal places)
03

Calculate the latent heat of fusion

We are given that the amount of energy required to melt the urethane is \(17.10 \mathrm{kJ}\). To find the latent heat of fusion in \(\mathrm{kJ/mol}\), we divide the energy required to melt the urethane by the number of moles. Latent heat of fusion, \(L = \frac{energy}{moles}\) \(L = \frac{17.10 \mathrm{kJ}}{1.136 \mathrm{mol}} = 15.05 \mathrm{kJ/mol}\) (rounded to 2 decimal places) The latent heat of fusion of urethane is \(15.05 \mathrm{kJ/mol}\).

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