Question

What is the molality of benzene in a solution that contains `"6.502 g"` of benzene in `"162.7 g"` of chloroform/`"CHCl"_3` (`K_f` for `"CHCl"_3` is `4.68^@ "C"cdot"kg/mol"` and `T_f^"*"` of `"CHCl"_3` is `63.5^@ "C"`)?

Answer 1

Molality is defined as moles of solute per `"1 kg = 1000 g"` of solvent. Convert benzene's mass to moles, then divide by the mass of the solvent in kilograms.

Explanation:

`"6.0502 g" * ("1 mol")/("78.114 g") = "0.083238 moles"`

So

`"molality" = "0.083238 moles"/(162.7 * 10^(-3) " kg") = "0.51156 mol kg"^(-1)`