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24 January, 08:49

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.)

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  1. 24 January, 09:16
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    The mole fraction of hexane is 0.67

    Explanation:

    This problem can be solved using Dalton's law and Raoult's law.

    The vapor pressure of a mixture of gases is the sum of the partial pressures of each gas (Dalton's law).

    Example:

    In a mixture of gases A and B

    Pt = PA + PB

    where:

    Pt = total pressure

    PA = partial pressure of A

    PB = partial pressure of B

    In an ideal solution, the vapor pressure of each component is equal to the vapor pressure of the pure component times the mole fraction of the component in the solution (Raoult's law).

    Example:

    In a solution containing A and B, the vapor pressure of A in the solution will be:

    PA = P (pure A) * Xa

    Where

    PA = vapor pressure of A in the solution

    P (pure A) = vapor pressure of pure A

    Xa = mole fraction of A

    In our problem, we have that the vapor pressure of the solution is 241 torr.

    Then, using Dalton's law:

    Pt = P (hexane) + P (Pentane)

    Using Raoult's law:

    P (hexane) = P (pure hexane) * X (hexane)

    P (pentane) = P (pure pentane) * X (pentane)

    We also know that the sum of the molar fractions of each component in a solution equals 1:

    X (hexane) + X (pentane) = 1

    X (pentane) = 1 - X (hexane)

    Replacing in the Dalton's law in terms of X (hexane):

    Pt = P (pure hexane) * X (hexane) + P (pure pentane) * (1 - X (hexane))

    Solving for X (hexane):

    Pt = P (pure hex) * X (hex) + P (pure pent) - P (pure pent) * X (hex)

    Replacing with the dа ta:

    241 torr = 151 torr * X (hex) + 425 torr - 425 torr*X (hex)

    -184 torr = - 274 torr * X (hex)

    -184torr/-274 torr = X (hex)

    X (hex) = 0.67
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