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RSB [31]
3 years ago
6

sample of substance X has a mass of 326.0 g releases 4325.8 cal when it freezes at its freezing point. If substance X has a mola

r mass of 58.45 g/mol, what is the molar heat of fusion for substance X?
Chemistry
2 answers:
Serhud [2]3 years ago
8 0

Answer:

C. 775.6 cal/mol

Explanation:

To put it straight forward

ELEN [110]3 years ago
5 0
Number of moles ( substance x ):

1 mole --------- 58.45 g/mol
? mole --------- 326.0 g

326.0 x 1 / 58.45 => 5.577 moles

 heat of fusion:

hf = Cal / moles

hf = 4325.8 Cal / 5.577 moles

hf = 775.65 cal/mol

hope this helps!
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This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and y_{C-6} and y_{N_2} are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

Q=-40036+(-672.21)=-40708.21J

Finally we convert this result to kJ:

Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ

Learn more:

  • brainly.com/question/25475410
  • brainly.com/question/12625048
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Answer:

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Explanation:

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