The normal boiling point of methanol is and the molar enthalpy of vaporization if 71.8 kJ/mol....

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The normal boiling point of methanol is and the molar enthalpy of vaporization if 71.8 kJ/mol....

The normal boiling point of methanol is and the molar enthalpy of vaporization if 71.8 kJ/mol. The value of S when 1.75 mol of vaporizes at 64.7 °C is 0.372 4.24 × 107 372 1.94 × 103 1.

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Answer 1

Answer #1

Q = number of mol * Hvap

= 1.75 mol * 71.8 KJ/mol

= 125.65 KJ

= 125650 J

T = 64.7 oC

= (64.7 + 273.1) K

= 337.8 K

Use:

S = Q/T

= 125650 J / 337.8 K

= 372 J/K

Answer: 372 J/K

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Answer 2

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The normal boiling point of methanol is and the molar enthalpy of vaporization if 71.8 kJ/mol....

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