It is found that 7.24 kJ of heat is required to vaporize 21.5 grams of liquid hexane, C6H14. Calculate the heat of vaporization of hexane in kJ/mol.
2007-05-25 07:09:20 · 4 answers · asked by aimeikins711 1 in Science & Mathematics ➔ Chemistry
7.24kJ ÷ 21.5g = 3kJ/g
1 mole of C6H14 has a mol.mass of (12x6) + (1x14) = 86.
This is expressed as 86g/mol.
86 ÷ 3 = 28.7kJ/mol = LH of vaporisation.
2007-05-25 07:58:49 · answer #1 · answered by Norrie 7 · 1⤊ 0⤋
Molar mass of C6H14 = 86 g
Thus Hv = 86*7.24/21.5 = 28.96 kJ/mol
2007-05-25 07:16:30 · answer #2 · answered by ag_iitkgp 7 · 2⤊ 0⤋
For heat of vaporization:
q= mass * (heat of vaporization)
q= total heat
7.24 kJ= 21.5g *(heat of vaporization)
You can do the rest.
2007-05-25 07:15:43 · answer #3 · answered by Jordan 3 · 0⤊ 1⤋
is this physical chemistry related or just general chemistry?
2007-05-25 07:37:42 · answer #4 · answered by wenjie x 1 · 0⤊ 0⤋