# What mass of water and urea were mixed, if the mol fraction of urea is 0.0825 and "60 g" of solution is present?

##### 2 Answers
Jan 8, 2017

The key is to know when to assume that the molar mass of the solution is the molar mass of the solvent, and when you can't.

That only works when the mol fraction of solute is really, really small (generally less than $0.010$ or so is a good ballpark), and is best when the molar masses of the solute and solvent are similar.

Otherwise, you still want to make sure conservation of mass still works, and that generally requires more work...

which would eventually give, for solute $i$ in solvent $j$:

${m}_{i} = \frac{{\chi}_{i} m {M}_{m , i}}{{\chi}_{j} {M}_{m , j} + {\chi}_{i} {M}_{m , i}}$

The definition of mol fraction ${\chi}_{i}$ of component $i$ says:

${\chi}_{i} = {n}_{i} / \left({n}_{1} + {n}_{2} + . . . + {n}_{N}\right)$

where $N$ is the number of components in the solution, and ${n}_{i}$ is the $\text{mol}$s of component $i$.

For a two component solution of urea and water, chi_"urea" = n_"urea"/(n_"urea" + n_"water").

Note that the relationship of mass $m$ to mols $n$ involves the molar mass ${M}_{m}$ like so:

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

So, if we divide the mass by the molar mass, we can convert the mass to mols for a substance we know the mass of in solution.

For your example:

chi_"urea" = 0.0825 = n_"urea"/(n_"urea" + n_"water")

${\chi}_{\text{water" = 1 - chi_"urea}} = 0.9175$

Unfortunately, since mol fractions are designed for mols and NOT masses, if we know the mass of the entire solution rather than the solvent only or the solute only, it does not make physical sense to use either the molar mass of urea or water to calculate the needed ${n}_{\text{tot}}$ (we instead need the "weighted-average" molar mass of the solution itself, which we don't have).

In other words, the mass percent of urea in solution is not equal to the mol fraction of urea in solution because ${M}_{m , \text{urea") ne M_(m,"water}}$.

SOLUTION BY ASSUMPTION

Provided the solution is sufficiently dilute (if ${\chi}_{\text{solute}}$ is small), we can say that the molar mass of the solution is approximately the molar mass of the solvent, i.e. ${M}_{m , \text{soln") ~~ M_(m,"solvent}}$.

What if ${\chi}_{\text{urea}} = 0.0025$? That's quite dilute. Then...

$0.0025 {n}_{\text{tot" = n_"urea" = "0.008325 mols urea}}$
$0.9975 {n}_{\text{tot" = n_"water" = "3.321675 mols water}}$

"0.008325 mols urea" xx "60.06 g urea"/"1 mol" = color(blue)("0.50 g urea")
"3.321675 mols water" xx "18.015 g water"/"1 mol" = color(blue)("59.84 g water")

That's only an apparent $\text{0.34 g}$ "increase" in the original mass of water, which is a 0.57% discrepancy with respect to conservation of mass --- not too bad!

Therefore, if ${\chi}_{\text{solute}}$ is really close to $0$, then we can assume that the molar mass of the solution is the molar mass of the solvent.

CONSIDERING CONSERVATION OF MASS AS AN ADDITIONAL EQUATION

You could also incorporate:

${m}_{\text{tot" = m_"water" + m_"urea}}$

so that ${m}_{\text{water" = m_"tot" - m_"urea}}$ $\text{g}$, and make this entirely general. Then, to keep this looking nice, let $i = \text{urea}$ and $j = \text{water}$, and $m = {m}_{\text{tot}}$, so that we have:

${n}_{i} = {m}_{i} / {M}_{m , i}$

${n}_{j} = \frac{m - {m}_{i}}{M} _ \left(m , j\right)$

From the definition of ${\chi}_{i}$:

$\implies {\chi}_{i} = \frac{{m}_{i} / {M}_{m , i}}{{m}_{i} / {M}_{m , i} + \frac{m - {m}_{i}}{M} _ \left(m , j\right)}$

= (m_i/M_(m,i))/((m_iM_(m,j) + (m - m_i)M_(m,i))/(M_(m,i)M_(m,j))

$= \left({m}_{i} / \cancel{{M}_{m , i}}\right) \cdot \frac{\cancel{{M}_{m , i}} {M}_{m , j}}{{m}_{i} {M}_{m , j} + \left(m - {m}_{i}\right) {M}_{m , i}}$

$= \frac{{m}_{i} {M}_{m , j}}{{m}_{i} {M}_{m , j} + \left(m - {m}_{i}\right) {M}_{m , i}}$

Or, solving for ${m}_{i}$:

${\chi}_{i} {m}_{i} {M}_{m , j} + {\chi}_{i} \left(m - {m}_{i}\right) {M}_{m , i} = {m}_{i} {M}_{m , j}$

${\chi}_{i} {m}_{i} {M}_{m , j} + {\chi}_{i} m {M}_{m , i} - {\chi}_{i} {m}_{i} {M}_{m , i} = {m}_{i} {M}_{m , j}$

${\chi}_{i} {m}_{i} {M}_{m , j} - {\chi}_{i} {m}_{i} {M}_{m , i} - {m}_{i} {M}_{m , j} = - {\chi}_{i} m {M}_{m , i}$

${m}_{i} \left[{\chi}_{i} {M}_{m , j} - {\chi}_{i} {M}_{m , i} - {M}_{m , j}\right] = - {\chi}_{i} m {M}_{m , i}$

Therefore:

$\textcolor{g r e e n}{{m}_{i}} = - \frac{{\chi}_{i} m {M}_{m , i}}{{\chi}_{i} {M}_{m , j} - {\chi}_{i} {M}_{m , i} - {M}_{m , j}}$

$= - \frac{{\chi}_{i} m {M}_{m , i}}{\cancel{{M}_{m , j}} - {\chi}_{j} {M}_{m , j} - {\chi}_{i} {M}_{m , i} - \cancel{{M}_{m , j}}}$

$= - \frac{{\chi}_{i} m {M}_{m , i}}{- {\chi}_{j} {M}_{m , j} - {\chi}_{i} {M}_{m , i}}$

$= \boldsymbol{\textcolor{g r e e n}{\frac{{\chi}_{i} m {M}_{m , i}}{{\chi}_{j} {M}_{m , j} + {\chi}_{i} {M}_{m , i}}}}$

Plugging in actual numbers, we'd get:

color(blue)(m_"urea") = (0.0825("60 g")("60.06 g/mol"))/((1-0.0825)("18.015 g/mol") + (0.0825)("60.06 g/mol"))

$=$ $\textcolor{b l u e}{\text{13.84 g urea}}$

which by conservation of mass gives you

$\textcolor{b l u e}{{m}_{\text{water") = "60.00 g" - "13.84 g urea" = color(blue)("46.16 g water}}}$

Jan 8, 2017

Given that total mass of aqueous solution of urea is $60 g$

Molar mass of urea${M}_{u} = \left(12 + 4 \cdot 1 + 2 \cdot 14 + 16\right) = 60 \frac{g}{\text{mol}}$

Molar mass of water${M}_{w} = 2 \cdot 1 + 1 \cdot 16 = 18 \frac{g}{\text{mol}}$

Let the mass of urea in the solution be ${m}_{u} = x g$

Then the mass of water in the solution ${m}_{w} = \left(60 - x\right) g$

The number of moles of urea in the solution ${n}_{u} = {m}_{u} / {M}_{u} = \frac{x g}{60 \frac{g}{\text{mol}}} = \frac{x}{60} m o l$

The number of moles of water in the solution ${n}_{w} = {m}_{w} / {M}_{w} = \frac{\left(60 - x\right) g}{18 \frac{g}{\text{mol}}} = \frac{60 - x}{18} m o l$

It is also given that mole fraction of urea in solution is

${\chi}_{u} = 0.0825$

By definition the mole fraction of urea in water-urea mixture is
${\chi}_{u} = {n}_{u} / \left({n}_{u} + {n}_{w}\right)$

Inserting the values in this relation we get

0.0825=(x/60)/(x/60+(60-x)/18

$\implies 0.0825 = \frac{18 x}{18 x + 3600 - 60 x}$

$\implies 18 x = 3600 \times 0.0825 - 42 \times 0.0825 x$

$\implies 18 x = 297 - 3.465 x$

$\implies 21.465 x = 297$

$\implies x = \frac{297}{21.465} = 13.836 g$

So $13.836 g$ urea is needed to dissolve in $\left(60 - 13.836\right) g = 46.164 g$ water to get an aqueous solution of urea where the mole fraction of urea is $0.0825$