Calculate the number of particles in each of the following:
$(a)$ $48 \,g$ of $Mg$
$(b)$ $8 \,g$ of $O_2$
$(c)$ $0.1$ mole of carbon (Atomic mass $Mg = 24 \,u, O = 16 \,u, C = 12 \,u$,Avogadro constant $= 6.022 \times 10^{23}$)
$(a)$ $48 \,g$ of $Mg$
$(b)$ $8 \,g$ of $O_2$
$(c)$ $0.1$ mole of carbon (Atomic mass $Mg = 24 \,u, O = 16 \,u, C = 12 \,u$,Avogadro constant $= 6.022 \times 10^{23}$)
(N/A) The number of atoms $= \frac{\text{Given mass}}{\text{Molar mass}} \times \text{Avogadro constant}$
$\Rightarrow N = \frac{m}{M} \times N_A = \frac{48}{24} \times 6.022 \times 10^{23} = 2 \times 6.022 \times 10^{23} = 12.044 \times 10^{23} \text{ atoms}$.
$(b)$ The number of molecules $= \frac{\text{Given mass}}{\text{Molar mass}} \times \text{Avogadro constant}$
$\Rightarrow N = \frac{8}{32} \times 6.022 \times 10^{23} = 0.25 \times 6.022 \times 10^{23} = 1.5055 \times 10^{23} \text{ molecules}$.
$(c)$ The number of particles (atoms) $= \text{Number of moles} \times \text{Avogadro constant}$
$\Rightarrow N = 0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22} \text{ atoms}$.
$\Rightarrow N = \frac{m}{M} \times N_A = \frac{48}{24} \times 6.022 \times 10^{23} = 2 \times 6.022 \times 10^{23} = 12.044 \times 10^{23} \text{ atoms}$.
$(b)$ The number of molecules $= \frac{\text{Given mass}}{\text{Molar mass}} \times \text{Avogadro constant}$
$\Rightarrow N = \frac{8}{32} \times 6.022 \times 10^{23} = 0.25 \times 6.022 \times 10^{23} = 1.5055 \times 10^{23} \text{ molecules}$.
$(c)$ The number of particles (atoms) $= \text{Number of moles} \times \text{Avogadro constant}$
$\Rightarrow N = 0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22} \text{ atoms}$.
Explore More
Similar Questions
Which of the following contains the maximum number of molecules?
Easy
View SolutionFind the total number of atoms present in one $SO_{4}^{2-}$ ion.
Easy
View SolutionComplete the following crossword puzzle by using the names of the chemical elements based on the data provided in the table.
Across Down $2$. The element used by Rutherford during his $\alpha$-scattering experiment. $1$. $A$ white lustrous metal used for making ornaments and which tends to get tarnished black in the presence of moist air. $3$. An element which forms rust on exposure to moist air. $4$. Both brass and bronze are alloys of this element. $5$. $A$ very reactive non-metal stored under water. $6$. The metal which exists in the liquid state at room temperature. $7$. Zinc metal when treated with dilute hydrochloric acid produces a gas of this element which when tested with a burning splinter produces a pop sound. $8$. An element with the symbol $Pb$.
| Across | Down |
|---|---|
| $2$. The element used by Rutherford during his $\alpha$-scattering experiment. | $1$. $A$ white lustrous metal used for making ornaments and which tends to get tarnished black in the presence of moist air. |
| $3$. An element which forms rust on exposure to moist air. | $4$. Both brass and bronze are alloys of this element. |
| $5$. $A$ very reactive non-metal stored under water. | $6$. The metal which exists in the liquid state at room temperature. |
| $7$. Zinc metal when treated with dilute hydrochloric acid produces a gas of this element which when tested with a burning splinter produces a pop sound. | $8$. An element with the symbol $Pb$. |
The atomic numbers of three elements $A, B$ and $C$ are $9, 10$ and $13$ respectively. Which of them will form a cation?
Medium
View Solution$(a)$ An atom is considered electrically neutral. Explain.
$(b)$ State the valency of Helium.
$(c)$ If an atom has an atomic mass of $4$ and $2$ protons in its nucleus,how many neutrons does it have?
$(b)$ State the valency of Helium.
$(c)$ If an atom has an atomic mass of $4$ and $2$ protons in its nucleus,how many neutrons does it have?
Medium
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