Question

Given that the vapor pressure of pure n-hexane and pure n-heptane at 25°c are 151.4 mmhg and 45.62 mmhg respectively, calculate the total vapor pressure above a solution containing only n-hexane and n-heptane in which the mole fraction of n-hexane is 0.600.

Answer 1

Answer is: total pressure is 108.96 mmHg.

p(n-hexane) = 151.4 mmHg; pressure of n-hexane.
p(n-heptane) = 45.62 mmHg.
χ(n-hexane) = 0.600; mole fraction.
χ(n-heptane) = 1 - 0.6 = 0.400.
Using Raoult's Law: 
p(total) = p(n-hexane) · χ(n-hexane)  + p(n-heptane) · χ(n-heptane).
p(total) = 151.4 mmHg · 0.6 + 45.32 mmHg · 0.4.
p(total) = 108.96 mmHg.

Answer 2

Total vapor pressure of the solution which is  containing only n-hexane and n-heptane is 108.96 mmHg at 25°C.

How we calculate the total vapor pressure?

Total vapor pressure of given solution will be calculated by using Raoult's Law as:

P = p(n-hexane) × χ(n-hexane)  + p(n-heptane) × χ(n-heptane), where

p(n-hexane) = vapor pressure of pure n-hexane = 151.4 mmHg

p(n-heptane) =vapor pressure of pure n-heptane = 45.62 mmHg

χ(n-hexane) = mole fraction of n-hexane = 0.600

χ(n-heptane) mole fraction of n-heptane = 1 - 0.600 = 0.400

Now putting all these value on the above equation, we get

P = (151.4 mmHg × 0.6) + (45.32 mmHg × 0.4) = 108.96 mmHg.

Hence, total vapor pressure of solution is 108.96 mmHg.

To know more about Raoult's Law, visit the below link:

https://brainly.com/question/12990462