Question #75537

1 Answer
Jul 4, 2016

Answer:

Here's what I got.

Explanation:

Your goal here is to find a way to go from liters of solution to kilograms of solvent by using the density of the solution.

As you know, a solution's molarity, #c#, tells you how many moles of solute you have in one liter of solution.

Let's assume that your starting solution has a molarity of #c# #"M"# and a density of #rho# #"g mL"^(-1)#. The first thing to do here is pick a sample of this initial solution.

Let's pick a #V# #"L"# sample. A solution that has a molarity of #c# #"M"# will contain #c# moles of solute in #"1 L"# of solution, which means that #V# #"L"# will contain

#V color(red)(cancel(color(black)("L solution"))) * (c color(white)(a)"moles solute")/(1color(red)(cancel(color(black)("L solution")))) = (c * V)color(white)(a)"moles solute"#

Now, the density of the solution is usually given in grams per milliliter, which means that you're going to have to convert the volume of the sample from liters to milliliters

#V color(red)(cancel(color(black)("L"))) * (10^3"mL")/(1color(red)(cancel(color(black)("L")))) = (V * 10^3)" mL"#

Use the density of the solution to find its mass

#(V * 10^3) color(red)(cancel(color(black)("mL solution"))) * (rhocolor(white)(a)"g")/(1color(red)(cancel(color(black)("mL solution")))) = (rho * V * 10^3)" g"#

You know that this sample contains #(c * V)# moles of solute. Use the molar mass of the solute, let's say #M_M# #"g mol"^(-1)#, to find how many grams of solute you have in this sample

#(c * V) color(red)(cancel(color(black)("moles solute"))) * (M_Mcolor(white)(a)"g")/(1color(red)(cancel(color(black)("mole solute")))) = (c * V * M_M)" g"#

Since the mass of the sample is made up of the mass of the solute and the mass of the solvent, you will have

#m_"solvent" = m_"solution" - m_"solute"#

This will get you

#m_"solvent" = (rho * V * 10^3)" g" - (c * V * M_M)" g" = V * (rho * 10^3 - c * M_M)" g"#

Now, the solution's molality, #b#, is calculated by taking the number of moles of solute present in one kilogram of solvent. Convert the mass of the solvent from grams to kilograms

#V * (rho * 10^3 - c * M_M) color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = (V * (rho * 10^3 - c * M_M))/10^3" kg"#

Since the sample contains #(c * V)# moles of solute in #(V * (rho * 10^3 - c * M_M))/10^3# kilograms of solvent, its molality will be

#b = (c color(red)(cancel(color(black)(V)))color(white)(a)"moles")/((color(red)(cancel(color(black)(V))) * (rho * 10^3 - c * M_M))/10^3"kg") = color(green)(|bar(ul(color(white)(a/a)color(black)((c * 10^3)/(rho * 10^3 - c * M_M) color(white)(a)"mol kg"^(-1))color(white)(a/a)|)))#

Here

#c# - the molarity of the solution in #"mol L"^(-1)#
#rho# - the density of the solution in #"g mol"^(-1)#
#M_M# - the molar mass of the solute in #"g mol"^(-1)#

An important thing to notice here is that the molality of the solution is independent of the volume of the sample, #V#, meaning that you have the same molality, and molarity, for that matter, regardless of the sample you pick.