Question #1f9b8
1 Answer
Explanation:
Your starting point here will be the definitions of molality,
Molality is defined as
#color(blue)(|bar(ul(color(white)(a/a)b = n/m_s ["mol"/"kg"]color(white)(a/a)|)))#
Here
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
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,
#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 = M_S/V#
Since the mass of the solution is equal to the mass of the solute,
#rho = (m + m_s)/V" " " "color(orange)((3))#
Use equation
#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
#c = m/(M_M * V) * 10^3 implies V = m/(M_M * c) * 10^3#
Use these two equations in equation
#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
#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
#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())#