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)|)))∣∣ ∣ ∣∣¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯aab=nms[molkg]aa∣∣∣−−−−−−−−−−−−−−−−−−
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)|)))∣∣ ∣ ∣∣¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯aab=nmS⋅10−3=nms⋅103[molg]aa∣∣∣−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Molarity is defined as
color(blue)(|bar(ul(color(white)(a/a)c = n/V["mol"/"L"]color(white)(a/a)|)))∣∣ ∣ ∣∣¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯aac=nV[molL]aa∣∣∣−−−−−−−−−−−−−−−−−
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)|)))∣∣ ∣ ∣∣¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯aac=nV⋅10−3=nV⋅103[molmL]aa∣∣∣−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
The number of moles of solute can be expressed using the mass of the solute,
n = m/M_Mn=mMM
Plug this into the equations for molality and molarity to find
b = m/(M_M * m_s) * 10^3" " " "color(orange)((1))b=mMM⋅ms⋅103 (1)
and
c = m/(M_M * V) * 10^3" " " "color(orange)((2))c=mMM⋅V⋅103 (2)
Now, you will need to use the density of the solution,
rho = M_S/Vρ=MSV
Since the mass of the solution is equal to the mass of the solute,
rho = (m + m_s)/V" " " "color(orange)((3))ρ=m+msV (3)
Use equation
b = m/(M_M * m_s) * 10^3 implies m_s = m/(M_M * b) * 10^3b=mMM⋅ms⋅103⇒ms=mMM⋅b⋅103
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)m+ms=m+mMM⋅b⋅103=m(1+1MM⋅b⋅103)
Use equation
c = m/(M_M * V) * 10^3 implies V = m/(M_M * c) * 10^3c=mMM⋅V⋅103⇒V=mMM⋅c⋅103
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)ρ=m⋅(1+1MM⋅b⋅103)m⋅1MM⋅c⋅103
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)ρ=MM⋅b+103MM⋅b⋅1MM⋅c⋅103
rho = (c * (M_M * b + 10^3))/(b * 10^3)ρ=c⋅(MM⋅b+103)b⋅103
All you have to do now is rearrange this to isolate
rho * b * 10^3 = c * M_M * b + c * 10^3ρ⋅b⋅103=c⋅MM⋅b+c⋅103
rho * b * 10^3 - c * M_M * b = c * 10^3ρ⋅b⋅103−c⋅MM⋅b=c⋅103
b(rho * 10^3 - c * M_M) = c * 10^3b(ρ⋅103−c⋅MM)=c⋅103
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)|)))∣∣ ∣ ∣∣¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯aab=c⋅103ρ⋅103−c⋅MMaa∣∣∣−−−−−−−−−−−−−−−−−−−−−−−
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)]ρ⋅103=[g mL−1]⋅103=[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)))))]b=[molL−1⋅103g⋅L−1−molL−1⋅gmol−1]
This has the units of
b = ["mol"/"g"] * 10^3b=[molg]⋅103
which is of course equivalent to
b = ["mol"/"kg"] " "color(green)(sqrt())b=[molkg] √