Question #75537
1 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,
Let's assume that your starting solution has a molarity of
Let's pick a
#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) 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,
#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
#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
An important thing to notice here is that the molality of the solution is independent of the volume of the sample,