You need to know the specific heat of vapourisation of benzene; ie; how many joules it takes to turn 1 mole of liquid benzene into 1 mole of gaseous benzene at the same temp.
Alternatively, if you know the entropy of liquid and gaseous benzene, you could work out the specific heat of vaporisation.
But given only the boiling point is not enough.
the link below provides the data you need; 30.77 kJ per mole.
Which means that one mole needs 30.77 kJ
Now all you need to work out is the number of moles in 396 grams.
formula mass of benzene is 78, so 1 mole is 78 grams.
which means that 396g is 5.077 moles
so you'll need 5.077 times as much as you need for one mole.
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There's not enough data to answer this question.
You need to know the specific heat of vapourisation of benzene; ie; how many joules it takes to turn 1 mole of liquid benzene into 1 mole of gaseous benzene at the same temp.
Alternatively, if you know the entropy of liquid and gaseous benzene, you could work out the specific heat of vaporisation.
But given only the boiling point is not enough.
the link below provides the data you need; 30.77 kJ per mole.
Which means that one mole needs 30.77 kJ
Now all you need to work out is the number of moles in 396 grams.
formula mass of benzene is 78, so 1 mole is 78 grams.
which means that 396g is 5.077 moles
so you'll need 5.077 times as much as you need for one mole.
I'm sure you can work that out.