Question

Calculate the molality and mole fraction of the solute is aqueous solution containing 3.0 gm of urea per 250 gm water (Mol.wt. of urea 60)?

Answer 1

Hint: We know the mathematical formula for Molality and Mole fraction. Use these
formulas to calculate. Here urea is the solute and water are the solvent.
.
Complete answer:
Molality is defined as the number of moles of solute dissolved in 1 kg of solute.
Its unit is mol/kg
Molal solution means 1 mole of solute is dissolved in 1kg of solution,its molality is one.
It is independent of temperature.
Formula for calculating molality,
\[Molality=\dfrac{Number\,of\,Moles}{Weight\,of\,sovent\,in\,Kg}\]
Mole fraction is defined as the ratio between number of moles of one component and total number of moles of all components present in solution.
It has no units and it is independent of temperature because volume is not involved in the formula.
\[Mole\,fraction\,=\,\dfrac{moles\,of\,subs\tan ce}{total\,number\,of\,moles}\]
The sum of mole fraction of all the components present in a solution is 1.
Now in the given question,
Mass of urea in the solution = 3g
Molar mass of urea = 60g/mol
Moles of urea = \[\dfrac{3g}{60g/mol}\] = 0.05
Mass of water = 250g
Molar mass of water = 18g/mol
Moles of water = \[\dfrac{250g}{18g/mol}\] = 13.88
Molality of the solute = \[\dfrac{Moles\,of\,solute}{Weight\,of\,sovent\,in\,Kg}\]
= \[\dfrac{0.05g}{250/1000}\]
= \[\dfrac{0.05}{0.250}\]
= 0.2 mol/kg
Mole fraction of solute = \[\dfrac{moles\,of\,solute}{total\,number\,of\,moles}\,\]
Let’s take a mole fraction of solute as \[{{X}_{solute}}\].
\[{{X}_{solute}}\] = \[\dfrac{0.05}{0.05\,+\,13.88}\]
\[{{X}_{solute}}\] = 0.0035
Thus, the molality is 0.2 mol/kg and mole fraction of solute is 0.0035.

Note: In contrast to molarity, which depends on the volume of the solution, molality depends only on the mass of the solvent. Therefore, it is independent of ambient changes such as temperature and pressure.