Tag: quantitative chemistry

Questions Related to quantitative chemistry

The critical volume of a gas is 0.036 $lit. mol^{-1}$. The radius of the molecule will be (in cm):
(Avogadro Number = $6 \times 10^{23}$)

chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

  1. $\displaystyle \left( \frac{9}{4 \pi} \times 10^{-23} \right)^{\dfrac{1}{3}}$

  2. $\displaystyle \left( \frac{8 \pi}{3} \times 10^{-23} \right)^{\dfrac{1}{3}}$

  3. $\displaystyle \left( \frac{3}{8 \pi} \times 10^{-23} \right)^{\dfrac{1}{3}}$

  4. none of these

Show answer
Correct Option: C
Explanation:

Critical volume of gas=3b [for 1 molecule]

For 1 mole, critical volume=0.036 liters
$\Longrightarrow 0.036\times { 10 }^{ 3 }{ cm }^{ 3 }=3\left( \cfrac { 4 }{ 3 } \pi { r }^{ 3 } \right) \times { N } _{ A } \ \Longrightarrow r={ \left( \cfrac { 36 }{ 24\pi  } \times 10^{ -23 } \right)  }^{ \cfrac { 1 }{ 3 }  }={ \left( \cfrac { 3 }{ 8\pi  } \times 10^{ -23 } \right)  }^{ \cfrac { 1 }{ 3 }  }$

The mass of a molecule of the compound $C {60}H _{122}$ is _________.

chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

  1. $1.4\times 10^{-21}$ g

  2. $1.09\times 10^{-21}$ g

  3. $5.025\times 10^{23}$ g

  4. $16.023\times 10^{23}$ g

Show answer
Correct Option: A
Explanation:

Molecular mass of $C _{60}H _{122}=(60 \times 12+1 \times 122)=720+122=842$


Hence, one mole contains $6.022 \times 10^{23}$ molecules 


Therefore, 

Mass of one molecule $=\frac{842}{6.022 \times 10^{23}}$

$=1.4 \times 10^{-21}\;g$


The correct option is A.

If you are given Avogadro's number of atoms of a gas $X$. If half of the atoms are converted into $X _{(g)}^+$ by energy $\Delta H$. The IE of $X$ is :

chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

  1. $\dfrac{2\Delta H}{N _A}$

  2. $\dfrac{2N _A}{\Delta H}$

  3. $\dfrac{\Delta H}{2N _A}$

  4. $\dfrac{N _A}{\Delta H}$

Show answer
Correct Option: A
Explanation:
Given no. of atoms = Avogadro's no. of atoms = ${N} _{A} = 6.023 \times {10}^{23}$
Given that $\cfrac{{N} _{A}}{2}$ atoms are ionized, i.e.,
Ionization energy of $\cfrac{{N} _{A}}{2}$ atoms of gas X = $\Delta{H}$

$\therefore$ Ionization energy of 1 atom of gas X = $\cfrac{\Delta{H}}{\left( \cfrac{{N} _{A}}{2} \right)} = \cfrac{2. \Delta{H}}{{N} _{A}}$
Hence, Ionisation energy of gas X is $\cfrac{2. \Delta{H}}{{N} _{A}}$.

Assuming that the all volume is measured at the same temperature and pressure, state the volume ratios of the reactants and products for the following gaseous reactions. Nitrogen reacting with oxygen to form nitrogen (III) oxide.

chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

  1. 1 : 3 : 2

  2. 2 : 4 : 2

  3. 2 : 2: 1

  4. 2 : 3 : 2

Show answer
Correct Option: D
Explanation:

The equation for the reaction between nitrogen and oxygen to form Nitrogen (lll) oxide is written as:
$2N _{2} + 3O _{2} \rightarrow 2N _{2}O _{3}$
Therefore the volume ratio of reactants and products $= 2 : 3 : 2$

Ethane burning in oxygen to give carbon dioxide and steam. The volume ratio of reactants to products is ______________.

chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

  1. 2 : 6 : 4 : 3

  2. 2 : 7 : 2 : 5

  3. 3 : 4 : 4 : 6

  4. 2 : 7 : 4 : 6

Show answer
Correct Option: D
Explanation:

The equation for the combustion reaction of ethane can be written as :
$2C _{2}H _{6} + 7O _{2} \rightarrow 4CO _{2} + 6H _{2}O$
Therefore the volume ratio of reactants and products $= 2 : 7 : 4 : 6$

The number of atoms in 67.2 L of ${ NH } _{ 3 }$(g) at STP is:

chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

  1. 9 ${N } _{ A }$

  2. 12 ${N } _{ A }$

  3. 3 ${N } _{ A }$

  4. 4 ${N } _{ A }$

Show answer
Correct Option: B
Explanation:
By Avogadro law,
 22.4 liter of any gas at STP$ = $1mole
                $ =6.022\times { 1 }0^{ 23 }$ molecules of gas
                $ =$NA (Avogadro number)

So, in 67.2 L of ${ NH } _{ 3 }=?$
    22.4 L of $ { NH } _{ 3 }=N _A$
     1 L of ${ NH } _{ 3 }=\cfrac { NA }{ 22.4 } $

67.2 L of $ { NH } _{ 3 }=\cfrac { 67.2 }{ 22.4 } \times NA$
                          $ =3N _A$ molecules

Now, each molecule contain 4 atoma
Number of atoms $= 4\times 3N _A = 12N _A$ atoms

Four one litre flasks are separately filled with gases $O _2, F _2, CH _4$ and $CO _2$ under same conditions. 


The ratio of the number of molecules in these gases are:

chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

  1. $2 : 2 : 4 : 3$

  2. $1 : 1 : 1 : 1$

  3. $1 : 2 : 3 : 4$

  4. $2 : 2 : 3 : 4$

Show answer
Correct Option: B
Explanation:
Avogadro's law: It states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

$\\ n _{1} : n _{2} : n _{3} : n _{4} = \cfrac{PV}{RT} : \cfrac{PV}{RT} : \cfrac{PV}{RT} : \cfrac{PV}{RT} = 1 : 1 : 1 : 1 $

So, the ratio of no. of molecules $= 1: 1: 1: 1$

The correct option is $B.$

A sample of municipal water contains one part of urea (molecular wt $=60$) per million parts of water by weight. The number of urea molecules in a drop of water of volume $0.05\ ml$ is 

chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

  1. $2.5\times 10^{14}$

  2. $5\times 10^{14}$

  3. $5\times 10^{13}$

  4. $5\times 10^{15}$

Show answer
Correct Option: B
Explanation:

$1 \ ppm = 1 \ mg/L = 10^{-3} g/L$

Water sample contains $1 \ ppm$ urea concentration.

$\therefore \ 10^{-3} \ g$ of urea in $ 1 \ L$

$x \ g$ urea in $0.05 \times 10^{-3} \ L$

$x= 0.05 \times 10^{-6} \ g$

$60 \ g$ urea $=6.023 \times 10^{23} \ molecules$

$0.05 \times 10^{-6} \ g = n \ molecules$

$n = \cfrac {6.023 \times 10^{23} \times 0.05 \times 10^{-6}}{60}$

$=0.005 \times 10^{17}$

$=5 \times 10^{14} \ molecules$

According to Avogadro's law the volume of a gas will ____ as _____ if ____ are held constant.

chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

  1. increases, number of moles; P & T

  2. decreases, number of moles; P & T

  3. increases; T & P; number of moles

  4. decreases; P & T; number of moles

Show answer
Correct Option: A
Explanation:

According to Avogadro's law: Equal volume of all gases at same temperature and pressure will have same no. of molecules.

OR
For a given mass of ideal gas,
Volume$\propto$Number of moles of the gas (if temperature and pressure are constant) 
So, volume of the gas will increase as the number of moles if pressure (P) and temperature (T) are held constant.

What is the value of $n$ in the following equation?


$Cr\left( OH \right) _{ 4 }^{ - }+OH^{ - }\longrightarrow  Cr{ O } _{ 4 }^{ 2- }+H _{ 2 }O\ +\ ne^-$

chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

  1. $3$

  2. $6$

  3. $5$

  4. $2$

Show answer
Correct Option: B