How many moles of $Mg_{3}(PO_{4})_{2}$ contain $0.25$ moles of $O$ atoms?
- A$2.55 \times 10^{-2}$
- B$0.0225$
- C$3.125 \times 10^{-2}$
- D$1.5 \times 10^{-2}$
Explore More
Similar Questions
What is the number of $CO_2$ molecules in $45.42 \ L$ at $\text{STP}$? (Consider molar volume of gas at $\text{STP} = 22.71 \ L \ mol^{-1}$)
The mass of one hydrogen atom is of the order of
Easy
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 SolutionWhat is the volume occupied by $16 \ g$ methane gas at $STP$ (in $cm^3$)?
What is the number of atoms present in $2.24 \ dm^3$ of $NH_3$ $(g)$ at $STP$?
Vedclass 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