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

Hint: Mole fraction is the ratio of moles of one component to the total number moles of all the components present in the solution.
$\text{mole}\text{fraction}\text{of}\text{solute (}{\text{X}}_B)\text{=}{\displaystyle \frac{\text{moles}\text{of}\text{solute (n)}}{\text{moles}\text{of}\text{solute (n)}\text{+}\text{moles}\text{of}\text{solvent (N)}}}.....(1)$

- 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
 $$\begin{array}{c}\text{mole}\text{=}{\displaystyle \frac{\text{mass}}{\text{molar}\text{mass}}}\\ N\text{=}{\displaystyle \frac{\text{1000}}{\text{18}}}\\ N\text{=}\text{55}\text{.55}\end{array}$$

To calculate the number of mole of solute we will apply equation no...(1)
 $$X_B={\displaystyle \frac{n}{n+N}}.....(1)$$
After putting the value of $$X_B$$and $$N$$in the equation (1) we get
 $$\begin{array}{rl}& X_B={\displaystyle \frac{n}{n+N}}\\ & 0.1={\displaystyle \frac{n}{n+55.55}}\\ & n=6.17\end{array}$$ $$\begin{array}{rl}& X_B={\displaystyle \frac{n}{n+N}}\\ & 0.1={\displaystyle \frac{n}{n+55.55}}\\ & n=6.17\end{array}$$
$$X_B={\displaystyle \frac{n}{n+N}}.....(1)$$
After calculating the no of mole of solute particle we will calculate the molality of the solution

$\begin{array}{rl}& \text{molality}\text{=}{\displaystyle \frac{\text{no}\text{of}\text{mole}\text{of}\text{solute(n)}}{\text{weight}\text{of}\text{solvent}\text{(Kg)}}}\\ & \text{molality}\text{=}{\displaystyle \frac{\text{6}\text{.17}}{1}}\\ So,& \text{molality}=6.17\end{array}$
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.