# Question 1f9b8

Jun 24, 2016

$b = \frac{c \cdot {10}^{3}}{\rho \cdot {10}^{3} - c \cdot {M}_{M}}$

#### Explanation:

Your starting point here will be the definitions of molality, $b$, and molarity, $c$.

Molality is defined as

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} b = \frac{n}{m} _ s \left[\text{mol"/"kg}\right] \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} b = \frac{n}{{m}_{S} \cdot {10}^{- 3}} = \frac{n}{m} _ s \cdot {10}^{3} \left[\text{mol"/"g}\right] \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Molarity is defined as

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} c = \frac{n}{V} \left[\text{mol"/"L}\right] \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} c = \frac{n}{V \cdot {10}^{- 3}} = \frac{n}{V} \cdot {10}^{3} \left[\text{mol"/"mL}\right] \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The number of moles of solute can be expressed using the mass of the solute, $m$, and its molar mass, ${M}_{M}$

$n = \frac{m}{M} _ M$

Plug this into the equations for molality and molarity to find

$b = \frac{m}{{M}_{M} \cdot {m}_{s}} \cdot {10}^{3} \text{ " " } \textcolor{\mathmr{and} a n \ge}{\left(1\right)}$

and

$c = \frac{m}{{M}_{M} \cdot V} \cdot {10}^{3} \text{ " " } \textcolor{\mathmr{and} a n \ge}{\left(2\right)}$

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

$\rho = \frac{m + {m}_{s}}{V} \text{ " " } \textcolor{\mathmr{and} a n \ge}{\left(3\right)}$

Use equation $\textcolor{\mathmr{and} a n \ge}{\left(1\right)}$ to find an expression for the mass of the solvent

$b = \frac{m}{{M}_{M} \cdot {m}_{s}} \cdot {10}^{3} \implies {m}_{s} = \frac{m}{{M}_{M} \cdot b} \cdot {10}^{3}$

The total mass of the solution will thus be

$m + {m}_{s} = m + \frac{m}{{M}_{M} \cdot b} \cdot {10}^{3} = m \left(1 + \frac{1}{{M}_{M} \cdot b} \cdot {10}^{3}\right)$

Use equation $\textcolor{\mathmr{and} a n \ge}{\left(2\right)}$ to find an expression for the volume of the solution

$c = \frac{m}{{M}_{M} \cdot V} \cdot {10}^{3} \implies V = \frac{m}{{M}_{M} \cdot c} \cdot {10}^{3}$

Use these two equations in equation $\textcolor{\mathmr{and} a n \ge}{\left(3\right)}$ to rewrite the density of the solution

$\rho = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{m}}} \cdot \left(1 + \frac{1}{{M}_{M} \cdot b} \cdot {10}^{3}\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{m}}} \cdot \frac{1}{{M}_{M} \cdot c} \cdot {10}^{3}}$

This will be equivalent to

$\rho = \frac{{M}_{M} \cdot b + {10}^{3}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{M}_{M}}}} \cdot b \cdot \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{M}_{M}}}} \cdot c} \cdot {10}^{3}}$

$\rho = \frac{c \cdot \left({M}_{M} \cdot b + {10}^{3}\right)}{b \cdot {10}^{3}}$

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

$\rho \cdot b \cdot {10}^{3} = c \cdot {M}_{M} \cdot b + c \cdot {10}^{3}$

$\rho \cdot b \cdot {10}^{3} - c \cdot {M}_{M} \cdot b = c \cdot {10}^{3}$

$b \left(\rho \cdot {10}^{3} - c \cdot {M}_{M}\right) = c \cdot {10}^{3}$

Therefore, the equation that establishes a relationship between molality and molarity is

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{b = \frac{c \cdot {10}^{3}}{\rho \cdot {10}^{3} - c \cdot {M}_{M}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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

For a molarity expressed in ${\text{mol L}}^{- 1}$, a density expressed in ${\text{g mL}}^{- 1}$, and a molar mass expressed in ${\text{g mol}}^{- 1}$, you will have

$\rho \cdot {10}^{3} = \left[{\text{g mL"^(-1)] * 10^3 = ["g L}}^{- 1}\right]$

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

$b = \left[\text{mol"/"g}\right] \cdot {10}^{3}$

which is of course equivalent to

b = ["mol"/"kg"] " "color(green)(sqrt())#