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20 April, 04:22

How many grams of water vapor (H2O) are in a 10.2 liter sample at 0.98 atmospheres and 26ÁC? Show all work used to solve this problem

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  1. 20 April, 04:36
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    The answer is 7.33 g.

    To calculate this, we will use the the ideal gas law:

    PV = nRT

    where

    P - pressure of the gas,

    V - volume of the gas,

    n - amount of substance of gas,

    R - gas constant,

    T - temperature of the gas.

    Since the amount of substance of gas (n) can be expressed as mass (m) divided by molar mass (M), then:

    PV = RTm/M

    It is given:

    P = 0.98 atm

    V = 10.2 l

    T = 26°C = 299.15 K

    R = 0.082 l atm/Kmol (gas constant)

    M (H2O) = 2Ar (H) + Ar (O) = 2*1 + 16 = 2 + 16 = 18g

    m = ?

    Since PV = RTm/M, then:

    m = PVM/RT

    m = 0.98 · 10.2 · 18 / 0.082 · 299.15 = 179.928/24.5303 = 7.33 g
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