The following information is given for ether, C2H5OC2H5, at 1atm:

boiling point = 34.6 °C Hvap(34.6 °C) = 26.5 kJ/mol
specific heat liquid = 2.32 J/g°C

At a pressure of 1 atm,
kJ of heat are needed to vaporize a 41.0 g sample of liquid ether at its normal boiling point of 34.6 °C.

I ASSUME the temperature of the liquid ether is at its boiling point of 34.6 C because you didn't list that; however, in problems of this type, especially since you gave the specific heat of the liquid, that the initial T is less than 34.6 C.

q needed to convert liquid ether at its boiling point to vapor at its boiling point is
q = mols ether x H vap = ?

To find the amount of heat needed to vaporize the sample, we can use the equation:

q = m * Hvap

Where:
q is the amount of heat required (in kJ)
m is the mass of the sample (in g)
Hvap is the heat of vaporization (in kJ/mol)

First, let's convert the given mass from grams to moles by using the molar mass of ether.

C: 12.01 g/mol
H: 1.01 g/mol
O: 16.00 g/mol

C2H5OC2H5 molar mass = (2 * C molar mass) + (5 * H molar mass) + (1 * O molar mass)

Now we can calculate the molar mass:

C2H5OC2H5 molar mass = (2 * 12.01 g/mol) + (5 * 1.01 g/mol) + (1 * 16.00 g/mol)

Next, divide the given mass of the sample by the molar mass to find the number of moles:

moles = mass / molar mass

Now that we have the number of moles, we can calculate the amount of heat required:

q = moles * Hvap

Finally, convert the amount of heat from J to kJ:

q(kJ) = q(J) / 1000

Now you can substitute the given values into the equation to find the amount of heat needed to vaporize the sample.