Tag: avogadro's law

Questions Related to avogadro's law

The volume of 2.8 g of $CO$ at $27^0C$ and 0.821 atm pressure is:

  1. 1.5 litre

  2. 3 litre

  3. 30 litre

  4. 0.3 litre


Correct Option: B
Explanation:

Molecular mass of $CO$ = 28. 

Moles of $CO$ present $= \dfrac{2.8}{28}$
 
$V=\dfrac{nRT}{P}=\dfrac{\dfrac{2.8}{28}\times0.0821\times300}{0.821}=3\ litres$

At what temperature will both Celsius and Fahrenheit scales read the same value:

  1. 100

  2. 180

  3. 40

  4. -40


Correct Option: D
Explanation:

We know, $\dfrac{F-180}{212-180}=\dfrac{C-273}{373-273}$


$F=C=-40$

So, the correct option is $D$

An isobaric process is associated with:

  1. constant temperature

  2. constant volume

  3. constant pressure

  4. small change in enthalpy


Correct Option: C
Explanation:

Isobaric process refers to a process taking place at constant pressure.

What could be better than a dozen $(12)$ donuts? How about a baker's dozen $(13)$ of donuts? Another large unit of measurement is known as Avogadro's number $(6.022\times 10^{23})$.
What is TRUE about Avogadro's number?

  1. Avogadro's number is the number of particles in one mole of any element.

  2. The molar mass of a substance will contain $6.022\times 10^{23}$ molecules.

  3. There are $6.022\times 10^{23}g$ of carbon in $12$ mol of $C - 12$.

  4. $6.022\times 10^{23}$ atoms of any element will have the exact same mass regardless of the identity of the element.


Correct Option: A
Explanation:

Avagadro constant i.e. $6.023 \times 10^{23}$ particles is the number of constituent particles, usually atoms or molecules, that are contained in the amount of substance given by $1\space mole$. 

In the ideal gas equation, the unit of universal gas constant R is?

  1. Mole-atm/K

  2. Litre/mole

  3. Litre-atm/mole.K

  4. erg/K


Correct Option: C
Explanation:

In ideal gas equation,


$R=\dfrac{PV}{nT}$ where, 
$P$ is the pressure in 'Atmosphere (atm)'
$V$ is the volume in Litre
$n$ is the number of moles
$T$ is the Temperature in Kelvin (K) 

$ \therefore $Units of $R$= litre.atm/mol.K

A sample of gas has a volume of 0.2 litres measured at 1 atm pressure and $0^oC$. At the same pressure, at $273^oC$, its volume will become:

  1. 0.1 litre

  2. 0.4 litre

  3. 0.8 litre

  4. 0.6 litre


Correct Option: B
Explanation:

According to Charles's law


$\dfrac{V _1}{T _1}=\dfrac{V _2}{T _2}$

$\dfrac{0.2}{273}=\dfrac{V}{546}$

$V=0.4\ litres$

Weight of 22.4 litres of $Cl _2$ gas at STP is:

  1. 17.25 g

  2. 35.5 g

  3. 73 g

  4. None


Correct Option: D
Explanation:

Volume occupied by 1 mole of any gas at STP is 22.4 litres. Therefore, in the question, we are given 1 mole of $Cl _2$ gas, which weigh 71 g.

A gas is initially at $1$ atm pressure. To compress it to $1/4$ th of its initial volume, pressure to be applied is:

  1. $1$ atm

  2. $2$ atm

  3. $4$ atm

  4. $1/4$ atm


Correct Option: C
Explanation:

According to boyle's law,


$P _1V _1=P _2V _2$

$\Rightarrow 1.V=\frac{V}{4}.P _2$

$\Rightarrow P _2=4\ atm$

Hence, the correct option is $C$

One mole of any substance contains $6.022\times 10^{23}$ atoms/molecules. Number of molecules of $H 2SO _4$ present in $100$mL of $0.02$M $H _2SO _4$ solution is _________?

  1. $12.044\times 10^{20}$ molecules

  2. $6.022\times 10^{23}$ molecules

  3. $1\times 10^{23}$ molecules

  4. $12.044\times 10^{23}$ molecules


Correct Option: A
Explanation:

Moles of $H _2SO _4=$ Molarity of $H _2SO _4\times $ Volume of solution $(L)$


$=0.02\times 0.1$


$=2\times 10^{-3}$ moles

No. of $H _2SO _4$ molecules $=2\times 10^{-3}\times 6.022\times 10^{23}=12.044\times 10^{20}$ molecules


Therefore, the correct option is A.

Cations absorb $6.023\times 10^{22}$ electrons for their reduction. How many equivalents of the ion are reduced?

  1. $0.1$

  2. $0.01$

  3. $0.001$

  4. $0.0001$


Correct Option: A
Explanation:

1 equivalent of ion absorbs 6. 023 $\times 10^{23}$ electrons 

No. of equivalents reduced = $\dfrac{6.023\times 10^{22}}{6.023 \times  10^{23}}$= 0.1