Question

QUESTION 1 1 points Save Answer The evaporation of ethanol (molar mass = 46.07 g/mol) involves an enthalpy change of 0.918 kJ/g. What is the molar enthalpy of vaporization of ethanol? 0 -0.0369 kJ/mol +37.8 kJ/mol O-42.3 kJ/mol O +1.18 kJ/mol O 12.2.1.1.al Click Save and Submit to save and submit. Click Save All Answers to save all answers. Save All Answers Close Window Save and Submit

Answer 1

Given : Ethanol (l) ( 1 g ) + 0.918 kJ $\rightarrow$ Ethanol ( vapor) ( 1 g)

We have relation, q = n $\Delta$ H vap. -------------------->(1)

Where, q is amount heat transferred to the substance,

n is no. of moles of substance ,

$\Delta$H vap. is enthalpy of vaporization of substance.

To calculate enthalpy of vaporization we need to calculate no. of moles of ethanol.

We have, No. of moles = Mass / Molar mass

$\therefore$ No. of moles of ethanol (n) = 1 g / ( 46.07 g /mol) = 0.021706 mol

Now, put q = 0.918 kJ , n = 0.021706 mol in equation 1, we get

0.918 kJ = 0.021706 mol ( $\Delta$ H vap )

$\therefore$ ( $\Delta$ H vap ) = 0.918 kJ / 0.021706 mol = 42.29 = 42.3 kJ / mol

ANSWER : $\Delta$ H vap of ethanol = 42.3 kJ / mol