Delta H Vap (h20)=40.7Kj/mol

How much water vaporized with 155Kj/mol?

I assume that is 155 kJ and not 155 kJ/mol.

40.7 kJ/mol x (?mol) = 155 kJ.

To find out how much water vaporized with 155 Kj/mol, we need to use the given enthalpy of vaporization (∆Hvap) for water, which is 40.7 Kj/mol.

The enthalpy change (∆H) is directly proportional to the amount of substance (in this case, moles of water vaporized). So, we can set up a proportion:

∆H1 / ∆H2 = n1 / n2

Where:
∆H1 = 40.7 Kj/mol (given ∆Hvap)
∆H2 = 155 Kj/mol (desired ∆Hvap)
n1 = unknown (moles of water vaporized corresponding to ∆H1)
n2 = unknown (moles of water vaporized corresponding to ∆H2)

Rearranging the equation, we can solve for n2:

n2 = (n1 * ∆H2) / ∆H1

Substituting the values:
n2 = (n1 * 155 Kj/mol) / 40.7 Kj/mol

Since n1 is unknown, we can substitute it with a variable, say 'x':

n2 = (x * 155 Kj/mol) / 40.7 Kj/mol

Now, we have an equation to determine the moles of water vaporized corresponding to ∆H2.

However, we don't have enough information to directly calculate the moles of water vaporized. We would need additional information like the temperature or pressure of the system to determine the number of moles of water vaporized. Without this information, it is not possible to find the exact amount of water vaporized corresponding to ∆H = 155 Kj/mol.