Chapter 13, Problem 13.138QP

(a) Derive the equation relating the molality (m) of a solution to its molarity (M)

$m=\frac{m}{d-\frac{MM}{1000}}$

where d is the density of the solution (g/mL) and m is the molar mass of the solute (g/mol). (Hint: Start by expressing the solvent in kilograms in terms of the difference between the mass of the solution and the mass of the solute.) (b) Show that, for dilute aqueous solutions, m is approximately equal to M.

Interpretation Introduction

Interpretation:

The equation relating the molality (m) of the solution to its molarity (M) has to be derived.

Concept introduction:

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent.

$\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Molarity (M): Molarity is number of moles of the solute present in the one liter of the solution.

$\text{Molarity (M) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1}\text{liter}\text{of}\text{solution}}$

Molarity is estimation of moles in the total volume of the solution while molality is estimation of moles in relationship with solvent in the solution

Answer to Problem 13.138QP

The equation relating the molality (m) of the solution to its molarity (M) is,

$\begin{array}{l}\text{kg solvent =}\left[\text{mass of the solution (g)- mass of the solute (g)}\right]\text{×}\frac{\text{1kg}}{\text{1000g}}\\ \text{or}\\ \text{kg solvent =}\frac{\text{(d)(1000) - M × molar mass}}{\text{1000}}\\ \text{or}\\ \frac{\text{m}}{\text{M}}\text{=}\frac{\text{1}}{\text{d-}\frac{\text{M×molar mass}}{\text{1000}}}\\ \text{or}\\ \text{m =}\frac{\text{M}}{\text{d-}\frac{\text{M×molar mass}}{\text{1000}}}\end{array}$

Explanation of Solution

Given data

$\text{m =}\frac{\text{M}}{\text{d-}\frac{\text{M}\text{. M'}}{\text{1000}}}$

Where,

d- Density of the solution ( $\text{g/mL}$)

M’- Molar mass of the solute (g/mol)

Relating molality(m) to molarity(M)

The equations relating the molality (m) of the solution to its molarity (M) are,

$\begin{array}{l}\text{kg solvent =}\left[\text{mass of the solution (g)- mass of the solute (g)}\right]\text{×}\frac{\text{1kg}}{\text{1000g}}\\ \end{array}$

Or

$\begin{array}{l}\text{kg solvent =}\frac{\text{(d)(1000) - M × molar mass}}{\text{1000}}\mathrm{......}(1)\\ \end{array}$

If we assume 1L of the solution then we can determine the mass of the solution from its volume and density and the mass of the solute from molar mass and molarity.

$\text{mass of the solution = d}\left(\frac{\text{g}}{\text{mL}}\right)\text{×1000mL}$

$\text{molar mass of solute = M}\left(\frac{\text{mol}}{\text{L}}\right)\text{×1L× molar mass}\left(\frac{\text{g}}{\text{mol}}\right)$

Substituting these expressions into equation (1) we get,

$\text{kg solvent =}\frac{\text{(d)(1000) - M × molar mass}}{\text{1000}}$

Or

$\text{kg solvent = d-}\frac{\text{M×molar mass}}{\text{1000}}\mathrm{......}\text{(2)}$

From molality definition we know that,

$\text{kg solvent =}\frac{\text{moles of the solute (n)}}{\text{m}}\mathrm{......}(3)$

We assume that 1L of the solution, we also know that mol solute (n) = Molarity (M), then the equation (3) becomes,

$\text{kg solvent =}\frac{\text{M}}{\text{m}}$

Substitute this expression into equation (2) we get,

$\frac{\text{M}}{\text{m}}\text{= d-}\frac{\text{M × molar mass}}{\text{m}}$

Take inverse of both sides of the equations,

$\frac{\text{m}}{\text{M}}\text{ = }\frac{\text{1}}{\text{d-}\frac{\text{M× molar mass}}{\text{1000}}}$

Or

$\text{m =}\frac{\text{M}}{\text{d-}\frac{\text{M×molar mass}}{\text{1000}}}$

Conclusion

The equation relating the molality (m) of the solution to its molarity (M) has been derived.

Interpretation Introduction

Interpretation:

For dilute aqueous solutions, m is approximately equal to M to be shown.

Concept introduction:

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent.

$\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Molarity (M): Molarity is number of moles of the solute present in the one liter of the solution.

$\text{Molarity (M) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1}\text{liter}\text{of}\text{solution}}$

Molarity is estimation of moles in the total volume of the solution while molality is estimation of moles in relationship with solvent in the solution

Answer to Problem 13.138QP

$\text{m}\approx \frac{\text{M}}{\text{d}}$

Explanation of Solution

 m is approximately equal to M For dilute aqueous solutions to be shown

For dilute aqueous solution the density is approximately $\text{1g/mL,}$ because water’ density is approximately $\text{1g/mL}\text{.}$

In dilute solutions, $\text{d >>}\frac{\text{M×molar mass}}{\text{1000}}$

For example, 0.010M $\text{NaCl}$ solution

$\frac{\text{M×molar mass}}{\text{1000}}\text{=}\frac{\left(\text{0}\text{.010mol/L}\right)\text{(58}\text{.44g/mol)}}{\text{1000}}\text{= 5}{\text{.8×10}}^{\text{-4}}\text{g/L<<1}$

$\text{with d >>}$ $\frac{\text{M×molar mass}}{\text{1000}}$

The derived equation reduces to

$\text{m}\approx \frac{\text{M}}{\text{d}}$

Because $\begin{array}{l}\text{d}\approx \text{1g/mL}\text{.}\\ \text{m}\approx \text{M}\end{array}$

Conclusion

For dilute aqueous solutions, m is approximately equal to M has been showed.