Question #b80d8

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.