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.