Question

Question #75537

Answer 1

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.