If a substance $A$ dissolves in a solution of a mixture of $B$ and $C$ with their respective number of moles as $n_A, n_B$ and $n_C$,the mole fraction of $C$ in the solution is:
- A$\frac{n_C}{n_A \times n_B \times n_C}$
- B$\frac{n_C}{n_A + n_B + n_C}$
- C$\frac{n_C}{n_A - n_B - n_C}$
- D$\frac{n_B}{n_A + n_B}$
Explore More
Similar Questions
The mole fraction of a solute in a $1$ molal ethanolic solution is:
Equal volumes of liquid $A$ $(d = 0.8 \ g/mL)$ and liquid $B$ $(d = 1.2 \ g/mL)$ are mixed to form a solution. Calculate the mole fraction of $A$ in the solution $(M_A = 16, M_B = 32)$.
Difficult
View SolutionMolarity of a solution prepared by dissolving $75.5 \ g$ of pure $KOH$ in $540 \ mL$ solution is .......... $M$
Medium
View SolutionWhat is the sum of the mole fractions of all the components in a solution?
Easy
View SolutionThe molality of the solution prepared by dissolving $125 \ mL$ of pure methanol $(d = 0.8 \ g \ mL^{-1})$ in $375 \ g$ of ethanol is
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo