Question

# The fluorocarbon compound$C_2Cl_3F_3$has a normal boiling point of$47.6 ^ { \circ } \mathrm { C }$The specific heats of$C_2Cl_3F_3 (l)$and$C_2Cl_3F_3(g)$are 0.91 and 0.67 J/g-K, respectively. The heat of vaporization for the compound is 27.49 kJ / mol. Calculate the heat required to convert 35.0 g of$C_2Cl_3F_3$from a liquid at$10.00 ^ { \circ } C$to a gas at$105.00 ^ { \circ } C$

Solutions

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The molar weights of $\text{C}$, $\text{Cl}$, and $\text{F}$ are $12.011\text{ g/mol}$, and $35.452\text{ g/mol}$, $18.998\text{ g/mol}$. Therefore, the molar weight of the molecule:

$\text{C}_2\text{Cl}_3\text{F}_3:12.011\times2+35.452\times3+18.998\times 3$

$\therefore \text{C}_2\text{Cl}_3\text{F}_3: 187.372\text{ g/mol}$

Find the number of moles of the molecule in the sample:

$\frac{50\text{ g}}{187.372\text{ g/mol}}=0.26685\text{ mols}$

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