Explain: Avogadro's Law.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) In $1811$,Italian scientist Amedeo Avogadro combined the conclusions of Dalton's atomic theory and Gay-Lussac's law of combining volumes,which is now known as Avogadro's Law.
Avogadro's Law: It states that equal volumes of all gases under the same conditions of temperature and pressure contain an equal number of molecules.
Mathematical Formula: According to Avogadro's Law,as temperature and pressure remain constant,the volume depends upon the number of molecules of the gas,or in other words,the amount of the gas.
$V \propto n$ (constant $T$ and $p$) .....(Eq.-$i$)
where,$n =$ number of moles of the gas.
$V = k_4 n$ (Eq.-$ii$)
Molar volume at $STP = 22.7 \ L$.
Atoms in one mole $= 6.022 \times 10^{23}$.
The relation between volume and density of a gas can be obtained as $M = k_4 d$.
Calculation of moles of gas:
Gaseous mole $(n)$ $= \frac{\text{Weight of gas (} m \text{)}}{\text{Molecular mass of gas (} M \text{)}}$ ....(Eq.-$i$)
$\therefore n = \frac{m}{M}$
where,$m =$ weight of gas,$M =$ molecular mass of gas.
According to Avogadro's law formula:
$V = k_4 n$ ....(Eq.-$ii$)
$\therefore V = k_4 \frac{m}{M}$ .....(Eq.-$iii$)
Thus,$M = k_4 \left( \frac{m}{V} \right)$ .....(Eq.-$iv$)
Since,$\frac{m}{V} =$ density of gas $= d$
$\therefore M = k_4 d$ (Eq.-$v$)
Conclusion: The density of a gas is directly proportional to its molecular mass.

Explore More

Similar Questions

Among the following,choose the ones with an equal number of atoms.
$A$. $212 \ g$ of $Na_2CO_3(s)$ [molar mass $= 106 \ g/mol$]
$B$. $248 \ g$ of $Na_2O(s)$ [molar mass $= 62 \ g/mol$]
$C$. $240 \ g$ of $NaOH(s)$ [molar mass $= 40 \ g/mol$]
$D$. $12 \ g$ of $H_2(g)$ [molar mass $= 2 \ g/mol$]
$E$. $220 \ g$ of $CO_2(g)$ [molar mass $= 44 \ g/mol$]
Choose the correct answer from the options given below.

The number of mole atoms and mole molecules in $56 \ L$ of oxygen gas at $STP$ are respectively:

$A$ flask contains $98 \ mg$ of $H_2SO_4$. If $3.01 \times 10^{20}$ molecules of $H_2SO_4$ are removed from the flask,the number of moles of $H_2SO_4$ remaining in the flask is $\left(N_A = 6.02 \times 10^{23}\right)$

The number of hydrogen atoms present in $5.4 \ g$ of urea is : (Given : Molar mass of urea: $60 \ g \ mol^{-1}$, $N_{A} = 6.022 \times 10^{23} \ particles \ mol^{-1}$)

$A$ sample of $[Cu(NH_3)_4]SO_4$ contains $2.4 \times 10^{24}$ ammonia molecules. The moles of $[Cu(NH_3)_4]SO_4$ in the given sample will be:

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