1.9 gm of a gas at 27 0C & 1.1 atm pressure occupies a volume of 1 ltr. The molecular mass of gas is, * 42.5 amu 3.83 amu

1.9 gm of a gas at 27 0C & 1.1 atm pressure occupies a volume of 1 ltr. The molecular mass of gas is, *
42.5 amu
3.83 amu
44 amu
32.5 amu

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  1. Answer:

    42.5 amu

    Explanation:

    Using the ideal gas law equation;

    PV = nRT

    P = pressure (atm)

    V = volume (L)

    n = number of moles (mol)

    R = gas law constant (0.0821 Latm/molK)

    T = temperature (K)

    According to the information provided in this question;

    mass = 1.9g

    T = 27°C = 27 + 273 = 300K

    P = 1.1atm

    V = 1L

    n = ?

    PV = nRT

    1.1 × 1 = n × 0.0821 × 300

    1.1 = 24.63n

    n = 1.1/24.63

    n = 0.045mol

    Mole = mass/molecular mass

    0.045 = 1.9/M.M

    M.M = 1.9/0.045

    M.M = 42.54amu

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