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

May 12, 2016

${\text{4.04 mol kg}}^{- 1}$

#### Explanation:

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

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{molality" = "moles of solute"/"kilogram of solvent} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, benzene, ${\text{C"_6"H}}_{6}$, is the solute and carbon tetrachloride, ${\text{CCl}}_{4}$, is the solvent, since there is more ${\text{CCl}}_{4}$, 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 $\text{12.0-g}$ 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

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 kg" = 10^3"g}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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.