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

1 Answer

Answer:

#"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.