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17 January, 20:03

What is the molality of a solution that contains 75.2 grams of AgClO4 in 885 grams of benzene? Question 5 options: 0.41 m 4.10 m 8.20 m 0.83 m

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  1. 17 January, 21:30
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    I took the test and got this question right! Here is the work.

    molality = moles solute / kg solvent

    moles AgClO4 = 75.2 g / 207.389 g/mol

    = 0.3626 mol

    molality = 0.6326 mol / 0.885 kg

    = 0.410 m

    The asnwer is 0.41.
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