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27 August, 16:27

If a student weighs out 0.744 g Fe (NO 3) 3 ⋅ 9 H 2 O, what is the final concentration of the ∼0.2 M Fe (NO 3) 3 solution that the student makes?

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  1. 27 August, 17:09
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    Molar concentration of Fe (NO3) 3. 9H2O = 0.12M

    Explanation:

    Fe (NO3).9H2O - -> Fe (NO3) 3 + 9H2O

    By stoichiometry,

    1 mole of Fe (NO3) 3 will be absorb water to form 1 mole of Fe (NO3) 3. 9H2O

    Therefore, calculating the mass concentration of Fe (NO3) 3;

    Molar mass of Fe (NO3) 3 = 56 + 3 * (14 + (16*3))

    = 242 g/mol

    Mass concentration of Fe (NO3) 3 = molar mass * molar concentration

    = 242 * 0.2

    = 48.4 g/L

    Molar mass of Fe (NO3) 3. 9H2O = 56 + 3 * (14 + (16*3)) + 9 * ((1*2) + 16)

    = 242 + 162 g/mol

    = 404g/mol

    Concentration of Fe (NO3) 3. 9H2O = mass concentration/molar mass

    = 48.4 / 404

    = 0.12 mol/l

    Molar concentration of Fe (NO3) 3. 9H2O = 0.12M
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