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23 December, 06:26

How many formula units are present in 473.7 g lead (ii) chloride?

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  1. 23 December, 07:56
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    First, divide the given mass of lead (ii) chloride by the molar mass of the compound. The molar mass is equal to 278.1 g/mol.

    n = m/M

    where n is the number of moles, m is the given mass, and M is the molar mass. Substituting the known values,

    n = 473.7 g / 278.1 g/mol

    n = 1.7 mols

    Then, it is to be remember that as per Avogadro, every mol of a substance contains 6.022 x 10^23 formula units.

    N = (1.7 mols) (6.022 x 10^23 formula units/mol)

    N = 1.02 x 10^24 formula units

    Answer: 1.02 x 10^24 formula units
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