A 0.50 g sample containing only anhydrous FeCl3 ( molar mass: 162.5gmol−1) and AlCl3 (molar mass: 133.5 g mol−1 yielded 1.435g of  AgCl (molar mass: 143.5gmol−1).  The mass of FeCl3 in the sample is

A 0.50 g sample containing only anhydrous FeCl3 ( molar mass: 162.5gmol1) and AlCl3 (molar mass: 133.5 g mol1 yielded 1.435g of  AgCl (molar mass: 143.5gmol1).  The mass of FeCl3 in the sample is

  1. A

    0.3126 g

  2. B

    0.4157 g

  3. C

    0.2345 g

  4. D

    0.1567 g

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    Solution:

    Amount of AgCl obtained, n=mM=1.435g143.5gmol1=0.01mol

    Let x be the mass of FeCl3, then

    Amount of FeCl3 =x162gmol1 and  Amount of AlCl3=0.50gx133.5gmol1

    Hence,  3x162gmol1+0.50gx133.5gmol1=0.01mol

    Solving for x, we get

    x=1(28.5)0.01×162×133.5381g=0.3126g

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