The number of sodium atoms in $2 \ mol$ of sodium ferrocyanide $Na_4[Fe(CN)_6]$ is .......... $\times 10^{23}$. (Given $N_A = 6.022 \times 10^{23} \ mol^{-1}$)
- A$12$
- B$26$
- C$34$
- D$48$
Explore More
Similar Questions
Numbers of gram atoms of oxygen present in $0.8 \ mol$ of $C_4H_6O_6$ would be
Difficult
View SolutionWhich one of the following has the maximum number of atoms?
MediumNEET 2020
View Solution$6.02 \times 10^{20}$ molecules of urea are present in $200 \ mL$ of its solution. The concentration of urea solution is .............. $M$
Difficult
View SolutionMatch List-$I$ with List-$II$. List-$I$ (Mass of substance) List-$II$ (Number of atoms) $A$. $1.8 \text{ mg}$ water $I$. $2 \times 10^{-4} \times N_{A}$ $B$. $9.8 \text{ mg}$ sulphuric acid $II$. $1.5 \times 10^{-4} \times N_{A}$ $C$. $1.8 \text{ mg}$ carbon $III$. $3 \times 10^{-4} \times N_{A}$ $D$. $5.85 \text{ mg}$ salt (NaCl) $IV$. $7 \times 10^{-4} \times N_{A}$
| List-$I$ (Mass of substance) | List-$II$ (Number of atoms) |
| $A$. $1.8 \text{ mg}$ water | $I$. $2 \times 10^{-4} \times N_{A}$ |
| $B$. $9.8 \text{ mg}$ sulphuric acid | $II$. $1.5 \times 10^{-4} \times N_{A}$ |
| $C$. $1.8 \text{ mg}$ carbon | $III$. $3 \times 10^{-4} \times N_{A}$ |
| $D$. $5.85 \text{ mg}$ salt (NaCl) | $IV$. $7 \times 10^{-4} \times N_{A}$ |
DifficultJEE MAIN 2026
View SolutionTwo flasks $A$ and $B$ of $500 \ mL$ each are filled with $O_2$ and $SO_2$ respectively at $300 \ K$ and $1 \ atm$ pressure. Then the flasks contain:
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