# 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

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 **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

#m_i = (chi_imM_(m,i))/(chi_jM_(m,j) + chi_iM_(m,i))#

The definition of **mol fraction**

#chi_i = n_i/(n_1 + n_2 + . . . + n_N)# where

#N# is the number of components in the solution, and#n_i# is the#"mol"# s of component#i# .

For a two component solution of urea and water,

Note that the relationship of mass

#M_m = m/n# , or#m = nM_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_"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

In other words, the mass percent of urea in solution is not equal to the mol fraction of urea in solution because

**SOLUTION BY ASSUMPTION**

Provided the solution is **sufficiently dilute** (if *approximately* the molar mass of the solvent, i.e.

What if

#0.0025n_"tot" = n_"urea" = "0.008325 mols urea"#

#0.9975n_"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

Therefore, if *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_"tot" = m_"water" + m_"urea"#

so that

#n_i = m_i/M_(m,i)#

#n_j = (m - m_i)/M_(m,j)#

From the definition of

#=> chi_i = (m_i/M_(m,i))/(m_i/M_(m,i) + (m - m_i)/M_(m,j))#

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

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

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

Or, solving for

#chi_im_iM_(m,j) + chi_i(m - m_i)M_(m,i) = m_iM_(m,j)#

#chi_im_iM_(m,j) + chi_imM_(m,i) - chi_im_iM_(m,i) = m_iM_(m,j)#

#chi_im_iM_(m,j) - chi_im_iM_(m,i) - m_iM_(m,j) = -chi_imM_(m,i)#

#m_i[chi_iM_(m,j) - chi_iM_(m,i) - M_(m,j)] = -chi_imM_(m,i)#

Therefore:

#color(green)(m_i) = -(chi_imM_(m,i))/(chi_iM_(m,j) - chi_iM_(m,i) - M_(m,j))#

#= -(chi_imM_(m,i))/(cancel(M_(m,j)) - chi_jM_(m,j) - chi_iM_(m,i) - cancel(M_(m,j)))#

#= -(chi_imM_(m,i))/(-chi_jM_(m,j) - chi_iM_(m,i))#

#= bb(color(green)((chi_imM_(m,i))/(chi_jM_(m,j) + chi_iM_(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"))#

#=# #color(blue)("13.84 g urea")#

which by conservation of mass gives you

#color(blue)(m_"water") = "60.00 g" - "13.84 g urea" = color(blue)("46.16 g water")#

Given that total mass of aqueous solution of urea is

Molar mass of urea

Molar mass of water

Let the mass of urea in the solution be

Then the mass of water in the solution

The number of moles of urea in the solution

The number of moles of water in the solution

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

By definition the mole fraction of urea in water-urea mixture is

Inserting the values in this relation we get

So