Question

Question #b80d8

Answer 1

Here's how you can determine the density of a solution if you know its molarity and its molality.

Molarity is defined as moles of solute per liters of solution.

`C = n/V_"solution"` `" "` `color(blue)((1))`

SIDE NOTE n will always represent the number of moles of solute, so I'll just leave it as n, instead of writing `n_"solute"`.

Since density is defined as mass per unit of volume, you can write the volume of the solution as

`rho = m/V => V_"solution" = m_"solution"/(rho)`

Use this in equation `color(blue)((1))` to get

`C = n/(m_"solution"/(rho)) = (n * rho)/m_"solution"` `" "color(blue)((2))`

You can write the mass of the solution as the sum of the mass of the solute and the mass of the solvent

`m_"solution" = m_"solute" + m_"solvent"`

Use this in equation `color(blue)((2))` to get

`C = (n * rho)/(m_"solute" + m_"solvent")` `" "color(blue)((3))`

Now use the solution's molality, which is defined as moles of solute per kilograms of solvent.

`b = n/m_"solvent" => m_"solvent" = n/b`

Use this in equation `color(blue)((3))` to get

`C = (n * rho)/((m_"solute" + n/b))` `" "color(blue)((4))`

Assuming you know what your solute is, you can write its mass as the product between the number of moles and its molar mass

`m_"solute" = n * M_M`

Use this in equation `color(blue)((4))` to get

`C = (n * rho)/((n * M_M + n/b)) = (cancel(n) * rho)/(cancel(n)(M_M + 1/b)) = (rho)/(M_M + 1/b)` `" "color(blue)((4))`

As a result, the solution's density can be expressed as

`color(green)(rho = C * (M_M + 1/b))`

Check to see if the units come out right

`rho = (cancel("moles"))/"L" * ("grams"/cancel("mole") + "kg"/cancel("mole"))`

`rho = "grams"/"L" + "kg"/"L"`

Convert the second term from kg per liter to grams per liter to get

`rho = "grams"/"L" + "grams"/"L" = "grams"/"L"`

And that's how you'd get the density of a solution from its molarity and molality.