A solution contains a mixture of pentane and hexane at room temperature. The solution has a vapor pressure of 241 torr. Pure pentane and hexane have vapor pressures of 425 torr and 151 torr, respectively, at room temperature.

What is the mole fraction of hexane? (Assume ideal behavior.)

To find the mole fraction of hexane in the solution, we need to use Raoult's law, which states that the partial pressure of each component in an ideal solution is proportional to its mole fraction multiplied by its vapor pressure at that temperature.

The vapor pressure of the solution is given as 241 torr, which is the sum of the partial pressures of pentane and hexane:

P_solution = P_pentane + P_hexane

Given that the vapor pressures of pentane (P_pentane) and hexane (P_hexane) are 425 torr and 151 torr, respectively, at room temperature, we can rewrite the equation as follows:

241 torr = (mole fraction of pentane) * 425 torr + (mole fraction of hexane) * 151 torr

Let's assume x represents the mole fraction of hexane in the solution. The mole fraction of pentane can be expressed as (1 - x) since the sum of the mole fractions must be 1.

Plugging these values into the equation, we have:

241 torr = (1 - x) * 425 torr + x * 151 torr

Simplifying the equation:

241 torr = 425 torr - 425x + 151x

Combining like terms:

241 torr - 425 torr = -274 torr = -274x

Dividing both sides by -274 torr:

x = 0.9993

Therefore, the mole fraction of hexane in the solution is approximately 0.9993.

write the missing number in each fraction so that the fraction belongs in the box. write one more fraction that can go in the box.

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If x is the mole fraction pentane, 1-x is the mole fraction of hexane.

Assume a mole-fraction-weighted vapor pressure for the mixture.

425x + 151(1-x) = 241

274 x = 90

Solve for x