## Related questions with answers

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

VerifiedThe 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:

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

-The boiling point of $\mathrm{C_2Cl_3F_3}$ is $T_b = 47.6 ^{\circ} \mathrm{C}$

-The specific heat of liquid $\mathrm{C_2Cl_3F_3}$ is $C_l = 0.91\mathrm{J/g.K}$

-The specific heat of gas is $C_g = 0.67 \mathrm{J/g.K}$

-The heat of vaporization of a compound is $\Delta H_{vap} = 27.49 \mathrm{kJ/mol}$

-The mass of $\mathrm{C_2Cl_3F_3}$ is $m = 35.0 \mathrm{g}$

Let us find how much heat is required to convert 35.0 $\mathrm{g}$ of liquid $\mathrm{C_2Cl_3F_3}$ at $10.00 ^{\circ} \mathrm{C}$ to a gas at $105.00 ^{\circ} \mathrm{C}$.

First, we will find the amount of heat needed to convert 35.0 $\mathrm{g}$ of liquid $\mathrm{C_2Cl_3F_3}$ at $10.00 ^{\circ} \mathrm{C}$ to $47.6 ^{\circ} \mathrm{C}$ of liquid $\mathrm{C_2Cl_3F_3}$.

$T_1 (l) = 10.00^{\circ} \mathrm{C} = (10.00 + 273) \mathrm{K} = 283 \mathrm{K}$

$T_2 (l) = 47.6^{\circ} \mathrm{C} = (47.6 + 273) \mathrm{K} = 320.6 \mathrm{K}$

The change in temperature is

$\Delta T (l) = T_2 (l) - T_1 (l) = 320.6 \mathrm{K} - 283 \mathrm{K} = 37.6 \mathrm{K}$

$\begin{align*} q_1 &= m \cdot C_l \cdot \Delta T (l)\\ &= 35.0 \mathrm{g} \cdot 0.91\mathrm{J/g.K} \cdot 37.6 \mathrm{K}\\ &= 1197.56 \mathrm{J} \end{align*}$

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