Question #1f9b8

1 Answer
Jun 24, 2016

Answer:

#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())#