if molar enthalpy of vaporization of enthanol is 38.6 kJ/mol, how many moles of ethanol are vaporize when required heat is 200.72kJ

q = moles x delta H vap.

solve for moles.

5.20 moL

Well, if the molar enthalpy of vaporization of ethanol is 38.6 kJ/mol, and we have 200.72 kJ of heat, we can use some math to figure it out.

Let's set up a proportion:

38.6 kJ / 1 mol = 200.72 kJ / x mol

Cross-multiplying, we get:

38.6 kJ * x mol = 1 mol * 200.72 kJ

Now we just have to solve for x:

38.6 kJ * x mol = 200.72 kJ

x mol = 200.72 kJ / 38.6 kJ

So, it looks like we need to borrow some mol-ecules from elsewhere, because we got x = 5.19 mol. Unfortunately, you can't really have a fraction of a mole, so I'm afraid you won't be able to vaporize exactly 200.72 kJ of heat with the given enthalpy of vaporization. Looks like you'll have to settle for 5 moles of ethanol vaporized, and hope the rest takes care of itself!

To determine the number of moles of ethanol vaporized when a certain amount of heat is transferred, we can use the equation:

Heat (in Joules) = moles * molar enthalpy of vaporization (in J/mol)

First, let's convert the given heat of 200.72 kJ to Joules:

200.72 kJ * (1000 J/1 kJ) = 200,720 J

Now, we rearrange the equation to solve for moles:

moles = Heat (in J) / molar enthalpy of vaporization (in J/mol)

moles = 200,720 J / 38,600 J/mol

Now, we can calculate the value:

moles ≈ 5.2 mol

Therefore, approximately 5.2 moles of ethanol will be vaporized when the required heat is 200.72 kJ.

5.20 mol is wrong