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11 November, 04:33

A 2.85-g sample of an unknown chlorofluorocarbon is decomposed and produces 564 mL of chlorine gas (Cl2) at a pressure of 752 mmHg and a temperature of 298 K. What is the percent chlorine (by mass) in the unknown chlorofluorocarbon

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  1. 11 November, 08:30
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    0.028%

    Explanation:

    A general formular for a chlorofluorocarbon (CFC) can be expressed as:

    R₂CFCL

    This is, an organic compund that has CL-C and F-C bonds (where R is a carbon chain)

    For the mentioned decomposition, reaction is:

    R₂CFCL → Cl₂ ...

    According to the information:

    P = 752 mm Hg

    T = 298 K

    V = 564 ml = 0.564 Lt of Cl₂

    If we consider that chlorine gas is an ideal gas, we may use the formula:

    PV=nRT = (m/M) RT

    m = (PVM) / (RT)

    M=molar mass of Cl₂ = 70 g/mol

    R = gas constant = 62.36 mm Hg*Lt/mol*K

    m = mass of Cl₂

    Then:

    m Cl₂ = (752x0.564x70) / (62.36x298)

    m Cl₂ = 0.0016 g

    If we consider (according to above reaction from CFC) that 2 moles of Cl₂ comes from 1 mol of Cl (from R₂CClF):

    mass of Cl on CFC = (0.0016) / 2 g = 0.0008 g

    So finally, the percent chlorine in the unknow chlorofluorocarbon is:

    % Cl = [ (0.0008) / 2.85) ]x100 = 0.0028%
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