Define the following terms:
$(i)$ Mole fraction $(ii)$ Molality $(iii)$ Molarity $(iv)$ Mass percentage

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) $(i)$ Mole Fraction: It is defined as the ratio of the number of moles of the solute to the total number of moles in the solution. If $n_A$ is the number of moles of solute and $n_B$ is the number of moles of solvent,then the mole fraction of solute $(X_A)$ is given by:
$X_A = \frac{n_A}{n_A + n_B}$
$(ii)$ Molality $(m)$: It is defined as the number of moles of solute present in $1 \ kg$ $(1000 \ g)$ of solvent.
$m = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}}$
$(iii)$ Molarity $(M)$: It is defined as the number of moles of solute present in $1 \ L$ of solution.
$M = \frac{\text{Number of moles of solute}}{\text{Volume of solution in litres}}$
$(iv)$ Mass Percentage: It is defined as the mass of solute in grams present in $100 \ g$ of solution.
$\text{Mass percentage} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$

Explore More

Similar Questions

When $6 \ g$ of urea is dissolved in $180 \ g$ of $H_2O$,the mole fraction of urea is:

The volumes of $4\,N\, HCl$ and $10\,N\, HCl$ required to make $1\, L$ of $6\,N\, HCl$ are

Difficult
View Solution

The hardness of a water sample due to $MgSO_4$ is $50 \, ppm$. Calculate the amount of $MgSO_4$ in $mg$ present in $1 \, kg$ of water.

Difficult
View Solution

What will be the mass percentage of an aqueous solution of $NaOH$ in which the mole fraction of $NaOH$ is $0.2$ (in $\% W/W$)?

If $5 \ g$ $NaOH$ is dissolved in $450 \ mL$ of solution,the molarity of the solution is: $(Na=23, O=16, H=1 \ g \ mol^{-1})$ (in $M$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo