What is the molality of a solution that contains 75.2 grams of agclo4 in 885 grams of benzene? a.) 0.41 m b.) 8.20 m c.) 4.10 m d.) 0.83 m

Answer is: C.) 4.10 m.
m(AgClO₄) = 75,2 g.
n(AgClO₄) = m(AgClO₄) ÷ M(AgClO₄).
n(AgClO₄) = 75,2 g ÷ 207,31 g/mol.
n(AgClO₄) = 0,362 mol.
m(C₆H₆) = 885 g ÷ 1000 g/kg = 0,885 kg.
b(solution) = n(AgClO₄) ÷ m(C₆H₆).
b(solution) = 0,362 mol ÷ 0,885 kg.
b(solution) = 0,41 mol/kg.

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