Question

Question #1f9b8

Answer 1

`b = (c * 10^3)/(rho * 10^3 - c * M_M)`

Explanation:

Your starting point here will be the definitions of molality, `b`, and molarity, `c`.

Molality is defined as

`color(blue)(|bar(ul(color(white)(a/a)b = n/m_s ["mol"/"kg"]color(white)(a/a)|)))`

Here

`n` - the number of moles of solute
`m_s` - the mass of the solvent expressed in kilograms

A more useful equation in this context has the mass of the solvent expressed in grams

`color(blue)(|bar(ul(color(white)(a/a)b = n/(m_S * 10^(-3)) = n/m_s * 10^3["mol"/"g"]color(white)(a/a)|)))`

Molarity is defined as

`color(blue)(|bar(ul(color(white)(a/a)c = n/V["mol"/"L"]color(white)(a/a)|)))`

Here

`n` - the number of moles of solute
`V` - the volume of the solution expressed in liters

Likewise, a more useful expression for this equation has the volume of the solution expressed in milliliters

`color(blue)(|bar(ul(color(white)(a/a)c = n/(V * 10^(-3)) = n/V * 10^3["mol"/"mL"]color(white)(a/a)|)))`

The number of moles of solute can be expressed using the mass of the solute, `m`, and its molar mass, `M_M`

`n = m/M_M`

Plug this into the equations for molality and molarity to find

`b = m/(M_M * m_s) * 10^3" " " "color(orange)((1))`

and

`c = m/(M_M * V) * 10^3" " " "color(orange)((2))`

Now, you will need to use the density of the solution, `rho`, as a way to convert between its mass, `M_S`, and its volume, `V`

`rho = M_S/V`

Since the mass of the solution is equal to the mass of the solute, `m`, plus the mass of the solvent, `m_s`, the equation becomes

`rho = (m + m_s)/V" " " "color(orange)((3))`

Use equation `color(orange)((1))` to find an expression for the mass of the solvent

`b = m/(M_M * m_s) * 10^3 implies m_s = m/(M_M * b) * 10^3`

The total mass of the solution will thus be

`m + m_s = m + m/(M_M * b) * 10^3 = m(1 + 1/(M_M * b) * 10^3)`

Use equation `color(orange)((2))` to find an expression for the volume of the solution

`c = m/(M_M * V) * 10^3 implies V = m/(M_M * c) * 10^3`

Use these two equations in equation `color(orange)((3))` to rewrite the density of the solution

`rho = (color(red)(cancel(color(black)(m))) * (1 + 1/(M_M * b) * 10^3))/(color(red)(cancel(color(black)(m))) * 1/(M_M * c) * 10^3)`

This will be equivalent to

`rho = (M_M * b + 10^3)/(color(red)(cancel(color(black)(M_M))) * b * 1/(color(red)(cancel(color(black)(M_M))) * c) * 10^3)`

`rho = (c * (M_M * b + 10^3))/(b * 10^3)`

All you have to do now is rearrange this to isolate `b` on one side of the equation

`rho * b * 10^3 = c * M_M * b + c * 10^3`

`rho * b * 10^3 - c * M_M * b = c * 10^3`

`b(rho * 10^3 - c * M_M) = c * 10^3`

Therefore, the equation that establishes a relationship between molality and molarity is

`color(green)(|bar(ul(color(white)(a/a)color(black)(b = (c * 10^3)/(rho * 10^3 - c * M_M))color(white)(a/a)|)))`

To make sure that the calculations are correct, test it out using units.

For a molarity expressed in `"mol L"^(-1)`, a density expressed in `"g mL"^(-1)`, and a molar mass expressed in `"g mol"^(-1)`, you will have

`rho * 10^3 = ["g mL"^(-1)] * 10^3 = ["g L"^(-1)]`

and

`b = [("mol"color(purple)(cancel(color(black)("L"^(-1)))) * 10^3)/("g" * color(purple)(cancel(color(black)("L"^(-1)))) - color(red)(cancel(color(black)("mol"))) color(purple)(cancel(color(black)("L"^(-1)))) * "g" color(red)(cancel(color(black)("mol"^(-1)))))]`

This has the units of

`b = ["mol"/"g"] * 10^3`

which is of course equivalent to

`b = ["mol"/"kg"] " "color(green)(sqrt())`