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30 January, 13:10

A volume of 18.0 l contains a mixture of 0.250 mole n2, 0.250 mole o2, and an unknown quantity of he. the temperature of the mixture is 0 ∘c, and the total pressure is 1.00 atm. how many grams of helium are present in the gas mixture?

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  1. 30 January, 16:10
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    Answer: - 1.22 g of helium are present in gas mixture.

    Solution: - Given, total pressure = 1.00 atm

    T = 0 + 273 = 273 K

    V = 18.0 L

    From ideal gas law equation, PV = nRT

    P = nRT/V

    Partial pressure of nitrogen = (0.250 x 0.0821 x 273) / 18.0

    partial pressure of nitrogen = 0.311 atm

    similarly, partial pressure of oxygen = (0.250 x 0.0821 x 273) / 18.0

    partial pressure of oxygen = 0.311 atm

    From Dalton's law of partial pressure, total pressure is the sum of partial pressures of the gases present in the container.

    So, total pressure = (partial pressure of nitrogen + partial pressure of oxygen + partial pressure of helium)

    partial pressure of helium = 1.00 atm - (0.311 atm + 0.311 atm)

    partial pressure of helium = 1.00 atm - 0.622 atm

    partial pressure of helium = 0.378 atm

    n = pV/RT

    So, moles of helium = (0.378 x 18.0) / (0.0821 x 273) = 0.304 mol

    Now we can calculate the grams of helium on multiplying it's moles by it's molar mass.

    0.304 mol x (4.00g/1mol) = 1.22 g
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