Question #1f9b8

1 Answer
Jun 24, 2016

b = (c * 10^3)/(rho * 10^3 - c * M_M)b=c103ρ103cMM

Explanation:

Your starting point here will be the definitions of molality, bb, and molarity, cc.

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

nn - the number of moles of solute
m_sms - 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)|)))∣ ∣ ∣¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯aab=nmS103=nms103[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

nn - the number of moles of solute
VV - 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)|)))∣ ∣ ∣¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯aac=nV103=nV103[molmL]aa−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

The number of moles of solute can be expressed using the mass of the solute, mm, and its molar mass, M_MMM

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=mMMms103 (1)

and

c = m/(M_M * V) * 10^3" " " "color(orange)((2))c=mMMV103 (2)

Now, you will need to use the density of the solution, rhoρ, as a way to convert between its mass, M_SMS, and its volume, VV

rho = M_S/Vρ=MSV

Since the mass of the solution is equal to the mass of the solute, mm, plus the mass of the solvent, m_sms, the equation becomes

rho = (m + m_s)/V" " " "color(orange)((3))ρ=m+msV (3)

Use equation color(orange)((1))(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) * 10^3b=mMMms103ms=mMMb103

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+mMMb103=m(1+1MMb103)

Use equation color(orange)((2))(2) to find an expression for the volume of the solution

c = m/(M_M * V) * 10^3 implies V = m/(M_M * c) * 10^3c=mMMV103V=mMMc103

Use these two equations in equation color(orange)((3))(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)ρ=m(1+1MMb103)m1MMc103

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=MMb+103MMb1MMc103

rho = (c * (M_M * b + 10^3))/(b * 10^3)ρ=c(MMb+103)b103

All you have to do now is rearrange this to isolate bb on one side of the equation

rho * b * 10^3 = c * M_M * b + c * 10^3ρb103=cMMb+c103

rho * b * 10^3 - c * M_M * b = c * 10^3ρb103cMMb=c103

b(rho * 10^3 - c * M_M) = c * 10^3b(ρ103cMM)=c103

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=c103ρ103cMMaa−−−−−−−−−−−−−−−−−−−−−

To make sure that the calculations are correct, test it out using units.

For a molarity expressed in "mol L"^(-1)mol L1, a density expressed in "g mL"^(-1)g mL1, and a molar mass expressed in "g mol"^(-1)g mol1, you will have

rho * 10^3 = ["g mL"^(-1)] * 10^3 = ["g L"^(-1)]ρ103=[g mL1]103=[g L1]

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=[molL1103gL1molL1gmol1]

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]