Question

For a solution with benzene as the solvent, what is the molality (with respect to the solute) if the solution has a solute mol fraction of `0.2`?

Answer 1

Since the mol fraction of solute is `chi_i = 0.2`, this indicates that we have:

`chi_i = 0.2 = (n_i)/(n_i+n_j)`

`= ("0.2 mols solute")/("0.2 mols solute" + "0.8 mols solvent")`

where `n_i` is the mols of `i`.

Recall that molality is:

`"mols solute"/"kg solvent"`

Fortunately, we know that the solvent is benzene, so the molar mass is:

`M_(j) = 6 xx 12.011 "g/mol C" + 6 xx 1.0079 "g/mol H"`

`=` `"78.1134 g/mol"`

This tells us that the molality is:

`color(blue)(m_"soln") = "0.2 mols solute"/(0.8 cancel"mols benzene") xx (cancel"1 mol benzene")/(78.1134 cancel"g benzene") xx (1000 cancel"g")/"1 kg"`

`=` `color(blue)("3.20 molal")`