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")?

Feb 15, 2018

Molality is defined as moles of solute per $\text{1 kg = 1000 g}$ of solvent. Convert benzene's mass to moles, then divide by the mass of the solvent in kilograms.
$\text{6.0502 g" * ("1 mol")/("78.114 g") = "0.083238 moles}$
${\text{molality" = "0.083238 moles"/(162.7 * 10^(-3) " kg") = "0.51156 mol kg}}^{- 1}$