Question 1 (1 point)
The heat of vaporization of methane, CH4, at its
boiling point is 9.20 kJ/mol. How much heat
energy is required to vaporize 100. g of methane
at its boiling point?

57.5 kj

Explanation:

Given data:

Heat of vaporization of  methane = 9.20 kj/mol

Heat energy required to vaporize 100 g of methane = ?

Solution:

First of all we will calculate the number of moles of methane in 100 g.

Number of moles of methane:

Number of moles = mass/ molar mass

Number  of moles = 100 g/ 16 g/mol

Number of moles = 6.25 mol

It is stated that for one mole of methane heat of vaporization is 9.20 kj.

For 6.25 mol or 100 g,

6.25 mol × 9.20 kj/mol

57.5 kj

try answer.com that works for me

i would say b

explanation:

but i think i would need more information than that