Question

If `"12.0 g"` of benzene is mixed with `"38.0 g"` of `"CCl"_4`, what is the molality?

Answer 1

`"4.04 mol kg"^(-1)`

Explanation:

A solution's molality tells you how many moles of solute you get per kilogram of solvent.

`color(blue)(|bar(ul(color(white)(a/a)"molality" = "moles of solute"/"kilogram of solvent"color(white)(a/a)|)))`

In your case, benzene, `"C"_6"H"_6`, is the solute and carbon tetrachloride, `"CCl"_4`, is the solvent, since there is more `"CCl"_4`, and both are liquids at STP.

The first thing to do here is use benzene's molar mass to determine how many moles you have in that `"12.0-g"` sample

`12.0 color(red)(cancel(color(black)("g"))) * ("1 mole C"_6"H"_6)/(78.112 color(red)(cancel(color(black)("g")))) = "0.1536 moles C"_6"H"_6`

The next thing to do here is convert the mass of the solvent from grams to kilograms by using

`color(purple)(|bar(ul(color(white)(a/a)color(black)("1 kg" = 10^3"g")color(white)(a/a)|)))`

You will have

`38.0 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = 38.0 * 10^(-3)"kg"`

The molality of the solution will thus be

`b = "0.1536 moles"/(38.0 * 10^(-3)"kg") = color(green)(|bar(ul(color(white)(a/a)"4.04 mol kg"^(-1)color(white)(a/a)|)))`

The answer is rounded to three sig figs.