Abottle is half-filled with a 50: 50 (mole-to-mole) mixture of n-heptane (c7h16) and n-octane (c8h18) at 25â°c. what is the mole ratio of heptane vapor to octane vapor in the air space above the liquid in the bottle? the vapor pressures of heptane and octane at 25â°c are 31 torr and 11 torr, respectively.

Answer is:  mole ratio of heptane vapor to octane vapor is 2,8 : 1.
p(C₇H₁₆) = 31 torr.
p(C₈H₁₈) = 11 torr.
x(C₇H₁₆) = 0,5.
x(C₈H₁₈) = 0,5.
p total = (x(C₇H₁₆) · p(C₇H₁₆)) + (x(C₈H₁₈) · p(C₈H₁₈)) = 0,5· 31 torr + 0,5·11 torr = 21 torr.
The mole fraction of heptane in the vapor is:
p(C₇H₁₆) ÷ p total) = 15,5 torr ÷ 21 torr = 0,74. 
The mole fraction of octane in the vapor is:
p(C₈H₁₈) ÷ p total = 5.5 torr ÷ 21 torr = 0,26.
Mole ratio heptane vapor ÷ octane vapor: 
0,74 ÷ 0,26 = 2,8.

Sorry i dont understand the question can you rephase it?