Question

The molarity of aq. Solution of urea in which mole fraction of urea is 0.1, is
(1) 6.17 m
(2) 5.24 m
(3) 6.98 m
(4) 5.78 m

Answer 1

Mole fraction is the ratio of moles of one component to the total number moles of all the components present in the solution.
$\text{mole}\,\,\text{fraction}\,\,\text{of}\,\text{solute (}{{\text{X}}_{B}}\,)\,\,\,\text{=}\dfrac{\text{moles}\,\text{of}\,\text{solute (n)}\,}{\,\text{moles}\,\text{of}\,\text{solute (n)}\,\text{+}\,\text{moles}\,\text{of}\,\text{solvent (N)}}.....(1)$

- Sum of mole fraction of solute and mole fraction of solvent is always one. ${{\text{X}}_{\text{A}}}\,\text{+}\,{{\text{X}}_{\text{B}}}\,\text{=}\,\,\text{1}$ Where, ${{\text{X}}_{\text{A}}}$ are the mole fraction of solvent and ${{\text{X}}_{\text{B}}}$the mole fraction of solute particles.

Complete Solution :
To calculate the molality of aq.solution first we will calculate the number of moles of solute after that we will calculate the molality of solution.
Mole fraction of solute (urea) is ${{\text{X}}_{B}}$ = 0.1
- Mole fraction of solute is 0.9. Molality is always calculated in 1000 gm and 1kg of solvent. So in the aqueous solution number of mole of solvent

\text{mole}\,\text{=}\dfrac{\text{mass}}{\text{molar}\,\text{mass}} \\\ N\,\,\text{=}\,\dfrac{\text{1000}}{\text{18}} \\\ \,\,N\,\,\text{=}\,\text{55}\text{.55} \\\ \end{matrix}$$ To calculate the number of mole of solute we will apply equation no...(1) $${{X}_{B}}\,=\,\,\dfrac{n}{n\,+\,N}.....(1)$$ After putting the value of $${{X}_{B}}\,$$and $$N$$in the equation (1) we get $$\begin{aligned} & {{X}_{B}}\,=\,\,\dfrac{n}{n\,+\,N} \\\ & 0.1\,\,=\,\dfrac{n}{n\,+\,55.55} \\\ & n\,\,\,=\,\,6.17 \\\ \end{aligned}$$ $$\begin{aligned} & {{X}_{B}}\,=\,\,\dfrac{n}{n\,+\,N} \\\ & 0.1\,\,=\,\dfrac{n}{n\,+\,55.55} \\\ & n\,\,\,=\,\,6.17 \\\ \end{aligned}$$ $${{X}_{B}}\,=\,\,\dfrac{n}{n\,+\,N}.....(1)$$ After calculating the no of mole of solute particle we will calculate the molality of the solution $\begin{aligned} & \text{molality}\,\,\text{=}\dfrac{\text{no}\,\text{of}\,\text{mole}\,\text{of}\,\text{solute(n)}}{\text{weight}\,\text{of}\,\text{solvent}\,\text{(Kg)}}\, \\\ & \text{molality}\,\,\text{=}\dfrac{\text{6}\text{.17}}{1}\, \\\ So, & \text{molality}\,\,=\,\,6.17 \\\ \end{aligned}$ **So, the correct answer is “Option 1”.** **Note:** molarity is defined as number of mole of solute particle is 1 liter volume of solution, however molality is number of mole of solute particle present in the 1kg weight of solvent. \- Mole fraction and molality are temperature independent while molarity is a temperature dependent quantity.