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24 August, 00:28

Calculate the density for camphor gas (C10H16O) at 12 Torr and 80 C.

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  1. 24 August, 00:42
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    0.084g/L

    Explanation:

    First, let us obtain an expression for the density. This is illustrated below:

    PV = nRT. (1)

    V = nRT/P (2)

    But n = Mass (m) / Molar Mass (M)

    n = m/M

    Substituting the value of n into equation 2

    V = nRT/P

    V = mRT/MP

    Divide both side by m

    V/m = RT/MP (3)

    Invert equation 3

    m/V = MP/RT (4)

    But Density = Mass (m) / volume (V)

    Density = m/V

    Replace m/V with Density in equation 4

    m/V = MP/RT

    Density = MP/RT

    The following data were obtained from the question:

    P = 12 Torr

    760torr = 1atm

    12torr = 12/760 = 0.016atm

    T = 80°C = 80 + 273 = 353K

    R = 0.082atm. L/K / mol

    Molar Mass (M) of C10H16O = (12x10) + (16x1) + 16 = 120 + 16 + 16 = 152g/mol

    Density = ?

    Density = MP/RT

    Density = (152 x 0.016) / (0.082x353)

    Density = 0.084g/L
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