Given below are the Mole Concepts Questions with Detailed Solutions
a. Concepts questions
b. Calculation problems
c. percentage composition
d. Mole fraction & molarity
Known Values
Question 1
How many molecules of water are there in 54 g of $H_2O$ ? Solution
Molar Mass of $H_2O$ = 2 + 16 = 18 g/moles
So ,number of moles of $H_2O$ = Mass/Molar Mass = 54/18 =3 moles
Now 1 moles = $6.022 \times 10^{23}$ molecule
So 3 moles will have $18.066 \times 10^{23}$ molecule
Question 2
Calculate the mass of $6.022 \times 10^{23}$ molecule of $NH_4Cl$ ? Solution
Molar mass (Molecular mass in gram) of $NH_4Cl$= 14+4+35.5= 53.5g
No. of moles of $NH_4Cl$
$=\frac { 6.022 \times 10^{23}}{ 6.022 \times 10^{23}}$
= 1 mole
Mass of $NH_4Cl$
$= \text {No. of moles} \times \text {molar mass}$
$= 1 \times 53.5g =53.5g$.
Question 3
Calculate the mass of $12.044 \times 10^{23}$ Oxygen atoms. Solution
No. of moles of Oxygen atoms
$= \frac {12.044 \times 10^{23}}{6.022 \times 10^{23}}$
= 2 mole
Mass of Oxygen atoms
$= \text {No. of moles} \times \text{atomic mass}$
$= 2 \times 16$
= 32 g
Question 4
How many atoms of hydrogen are there in 36 g of $NH_4$? Solution
Molar mass (Molecular mass in gram) of $NH_4$= 14+4= 18 g
No. of moles of $NH_4$?
= 36/18 = 4 moles
Now Total Moles of Hydrogen Atoms
= 16 moles
$= 16 x 6.022 \times 10^{23}=96.352 \times 10^{23}$
Question 5
Calculate the number of hydrogen atoms in 1 mole of H2. Solution
1 molecule of H2= 2 hydrogen atoms
So, 1 mole of H2= 2 mole hydrogen atoms
$= 2 \times 6.022 \times 10^{23}=12.044 \times 10^{23}$ hydrogen atoms.
Question 6
Calculate the number of Cu atoms in 0.3175 g of Cu. Solution
No. of moles of Cu
$= \frac {\text {Mass of Cu}}{\text {Atomic mass}}$
= 0.3175/63.5
=0.005 mole
No. of Cu atoms
$= \text {No. of moles} \times \text {Avogadro constant}$
$= 0.005 \times 6.022 \times 10^{23}$
$= 30.11 \times 10^{20}$ Cu atoms.
Question 7
Calculate the number of molecules in 22.4 litres of CH4 gas at NTP. Solution
1 mole of CH4= 22.4 L (at NTP)
Therefore
22.4 L of CH4gas= 1 mole CH4gas= $6.022 \times 10^{23}$ CH4 gas molecules.
Question 8
a. Find the volume of 1 g of H2gas in litres at N.T.P
b.Find the volume of 20g H2at NTP.
c.What is the volume occupied by $6.022 \times 10^{23}$ molecules of any gas at NTP? Solution
a. Molar Mass of H2gas = 2 gm
$\text {Number of moles} = \frac {Mass}{\text {Molar Mass}} = \frac {1}{2} = .5$
1 mole of H2= 22.4 L (at NTP)
Therefore
.5 mole of H2= .5 x 22.4 = 11.2 litre
b.
No. of moles of H2= 20/2 =10
1 mole of H2= 22.4 L (at NTP)
Therefore
10 moles = $10 \times 22.4$
=224 L
c.
$6.022 \times 10^{23}$ molecules = 1 mole molecules, and
1 mole molecules of any ideal gas occupies 22.4 L at NTP.
Question 9
An atom of some element Y weighs $6.644 \times 10^{-23}$ g. Calculate the number of gram-atoms in 40 kg of it. Solution
Mass of 1 mole Y atoms
$= \text {mass of 1 atom} \times \text {Avogadro constant}$
$= 6.644 \times 10^{-23} \times 6.022 \times 10^{23}$
= 40 g
So, the atomic mass of Y = 40
No. of gram-atoms (or moles) of X
$ = \frac {\text {mass of Y}}{\text {atomic mass}}$
$= \frac {40 \times 1000}{40} =1000$
Question 10
Find the number of moles and number of atoms of H and S in 10 mole of $H_2S$. Solution
1 mole of H2S contains 2 mole of H, 1 mole of S
Therefore
10 mole of H2S contains
20 mole of H = $20 \times 6.022 \times 10^{23}= 12.044 \times 10^{24}$ H atoms
10 mole of S = $10 \times 6.022 \times 10^{23}= 6.022 \times 10^{24}$ S atoms
Question 11
Calculate the number of atoms of each element in 245 g of $KClO_3$. Solution
Molecular mass of KClO3= $39 +35.5+3 \times 16 = 122.5$
No. of mole of KClO3= 245 g/122.5g = 2 mole
2 mole of KClO3contains
2 mole of K = $2 \times 6.022 \times 10^{23} = 12.044 \times 10^{23}$ K atoms
2 mole of Cl = $2 \times 6.022 \times 10^{23} = 12.044 \times 10^{23}$ Cl atoms
6 mole of O = $6 \times 6.022 \times 10^{23}= 1.806 \times 10^{24}$ O atoms.
Question 12
If 4 g of NaOH dissolves in 36 g of $H_2O$, calculate the Mole fraction of each component in the solution. Also, determine the Molarity of solution (specific gravity of solution is 1 gm/L) Solution
Mass of NaOH = 4 g
Number of moles of = 4 g NaOH/40 g = 0.1 mol
Mass of $H_2O$ = 36 g
Number of moles of = 36 g/18= 2 mol
$\text {Mole fraction of water} =\frac {\text{Number of moles of water}} {\text{ No. of moles of water} + \text {No. of moles of NaOH}}$
$ = \frac {2}{2 +.1}= 0.95$
$\text {Mole fraction of NaOH} = \frac {\text {Number of moles of NaOH}}{ \text {No. of moles of NaOH} +\text { No. of moles of water}}$
$ = \frac {0.1}{ 2 + 0.1} = 0.047$
Mass of solution = mass of water + mass of NaOH = 36 g + 4 g = 40 g
Volume of solution = $40 \times 1 = 40$ mL (Since specific gravity of solution is = 1 g m/L)
$\text {Molarity of solution}= \frac {\text {Number of moles of solute }}{\text {Volume of solution in litre}} =\frac {0.1}{0.04} =2.5$ M
Question 13
Calculate the number of atoms in 12.2 litres of the below gas at NTP
(i) mono-atomic
(ii) diatomic gas Solution
No. of moles in 12.2 L gas at NTP = 12.2/22.4 = 0.50
No. of molecules in 12.2 L gas
$= 0.50 \times 6.022 \times 10^{23}$
$= 3.011 \times 10^{23}$ molecules
In mono-atomic gases, No. of atoms = No. of molecules =$3.011 \times 10^{23}$.
In diatomic gases, No. of atoms = $2 \times \text{No. of molecules} =2 \times 3.011 \times 10^{23}=6.022 \times 10^{23}$.
Question 14
If a mole were to contain $1 \times 10^{24}$ particles, what would be the mass of
(i) one mole of oxygen
(ii) a single oxygen molecule? Solution
Mass of one mole of oxygen molecule (O2)
= molecular mass of oxygen molecule (O2) in gram
= 32 g.
Mass of a single oxygen molecule
$= \frac {32}{1 \times 10^{24}}$
$=3.2 \times 10^{-23}$ g
Question 15
Calculate the standard molar volume of oxygen gas. The density of $O_2$ gas at NTP is 1.429 g/L. Solution
Standard molar volume is Volume of 1 mole of gas i.e 32 gm O2 gas
Now
$Density = \frac {Mass}{volume}$
$Volume= \frac {Mass}{Density}$
= 32/1.429
=22.39 Litres.
Question 16
if 2 mol of Calcium Carbonate (Formula Weight =100) occupies a volume of 67.0 ml, Find the density ? Solution
Mass of Calcium Carbonate = 2 * 100 = 200 g
Volume = 67.0 ml
$Density = \frac {Mass}{volume} = \frac {200}{67}= 2.99$ g/mL
Question 17
Calculate how many methane molecules and how many carbon and hydrogen atoms are there in 25 g of Methane? Solution
Molar mass pf methane =16
Number of Moles =25/16
No of methane molecules = $\frac {25}{16} \times 6.022 \times 10^{23}= 9.411 \times 10^{23}$
No of carbon molecules = $1 \times 9.411 \times 10^{23}= 9.411 \times 10^{23}$
No of hydrogen molecules = $4 \times 9.411 \times 10^{23}= 3.74 \times 10^{24}$
Question 18
A metal M of atomic mass
54.94 has a density of 7.42g/cc. Calculate the apparent volume occupied by one atom of the metal. Solution
Mass of 1 mole metal atoms = 54.94 g
Mass of 1 metal atom = $ \frac {54.94}{6.022 \times 10^{23}} = 9.12 \times 10^{23}$g
Volume occupied by one metal atom
$= \frac {\text{Mass of one metal atom}}{density}$
$= \frac {9.12 \times 10^{23}}{7.42}$
$=1.23 \times 10^{23}$ cc.
Question 19
Calculate the number of moles of Ca, C and oxygen atoms and its mass in 200 g of $CaCO_3$. Solution
Molecular mass of CaCO3= 40+12+3 \times 16 = 100
No. mole of CaCO3= $ \frac {200 g}{100 g} = 2$
Number of Moles of Ca=2 moles
Number of Moles of C=2 moles
Number of Moles of O=6 moles
Question 20
Calculate the total number of electrons present in 3.2 g of CH4. Solution
Molecular mass of CH4= $12+4 \times 1 = 16$
Moles of CH4=$\frac {3.2}{16} =0.2$
No. of electron in 1 molecule of CH4= 6+4 = 10 electrons
Total no. of electrons = $0.2 \times 6.022 \times 10^{23} \times 10 =12.044 \times 10^{23}$ electrons.
Question 21
How much Calcium is in the amount of Ca(NO3)2 that contains 20.0g of Nitrogen? Solution
Number of Moles of Nitrogen = $ \frac {20}{14} =1.428$ moles
From the Molecular formula, it is apparent that, the number of Moles Ca will be half number of Number of Nitrogen
So, number of Moles of Ca=$ \frac {1.428}{2} = .714$ moles
So mass of Ca = $.714 \times 40 = 28.56$ g
Question 22
State the number of atoms in 1 g atom of Aluminium? Solution
1 gm atom = atomic weight
So number of atoms will be Avogadro number = $6.022 \times 10^{23} $
Question 23
If the components of the air are N2, 78%; O2, 21%; Ar, 0.9% and CO2, 0.1% by volume, what would be the molecular mass of air? Solution
The molar ratios are also volume ratios for gases (Avogadro’s principle)
Molecular mass of air
$= \frac {78 \times 28 + 21 \times 32 + 0.9 \times 40 + 0.1 \times 44}{78 + 21 + 0.9 + 0.1}$
=28.964.
Question 24
How many molecules are there in 1.624 gm Ferric chloride(FeCl3)? Solution
Molar Mass of Ferric chloride= 56 + 3 * 35.5 = 162.5 gm
Number Of moles = 1.624 /162.5
Number Of molecules = $(\frac {1.624}{162.5} ) \times 6.022 \times 10^{23}= 6.023 \times 10^{21}$ Molecules
Question 25
What is the mass of 1 Ammonia Molecule ? Solution
Molar Mass of Ammonia= $14 + 3 \times 1= 17$ gm
Mass of One molecule =$ \frac { 17}{(6.022 \times 10^{23})} = 2.8 \times 10^{-23}$ g
Question 26
The atomic masses of two elements (P and Q) are 20 and 40 respectively. x g of P contains y atoms, how many atoms are present in 2x g of Q? Solution
No. of mole of P = $\frac {x}{20}$
No. of atoms of P = $(\frac {x}{20}) \times N$ [N is Avogadro constant]
Therefore, $y = x \times \frac {N}{20}$
or $x= \frac {20y}{N}$
Now,
No. of mole of Q = $\frac {2x}{40}$
No. of atoms of Q
$= (\frac {2x}{40}) \times N$
$= \frac {2N}{40} \times \frac {20y}{N}$
=y
Question 27
Oxygen is present in a 1-litre flask at a pressure of $7.6 \times 10^{-10}$ mm of Hg at 0°C. Calculate the number of oxygen molecules in the flask. Solution
Pressure = $7.6 \times 10^{-10}$ mm Hg
as 1 atm = 760 mm Hg
$p= \frac {7.6 \times 10^{-10}}{ 760}$
$= 10^{-12}$ atm
Volume = 1 litre
Temperature = 0°C = 273 K
We know $pV = nRT$
or
$n = \frac {pV}{RT}$
$n = \frac {(10^{-12} \times 1)}{(0.0821 \times 273)} = 0.44 \times 10^{-13}$
$\text {No. of molecules} = \text {no. of moles} \times \text {Avogadro constant}$
$= 0.44 \times 10^{-13} \times 6.022 \times 10^{23}$
$=2.65 \times 10^{10}$ .
Question 28
What is the ratio of the volumes occupied by 1 mole of O2and 1 mole of O3in identical conditions? Solution
Volume ratio
= Molar ratio (Avogadro’s principle – the molar ratios are also volume ratios for gases)
=1:1
Question 29
Calculate the mass of .5 moles of CaCO3in g. Solution
Molar mass (i.e., molecular mass in g) = $40+12+3 \times 16 = 100$g
Mass of .5 moles of CaCO3= $.5 \times 100 = 50$ g
Question 30
The cost of the Table Salt(NaCl) and Table sugar($C_{12}H_{22}O_{11}$) is Rs 10 and Rs 40 per kg. Find the cost of the salt and sugar per mole? Solution
Molar Mass of NaCl= $23 + 35.5 = 58.5$ g/mol
Cost of 1 Mole of NaCl =$ (\frac {10}{1000} ) \times 58.5 = .585$ Rs/mole
Molar Mass of $C_{12}H_{22}O_{11}$= $12 \times 12 + 22 \times 1 + 11 \times 16 = 342$ g/mol
Cost of 1 Mole of $C_{12}H_{22}O_{11}$ = $( \frac {40}{1000} ) \times 342 = 13.68$ Rs/mole
Question 31
Calculate the number of oxygen atoms in 0.2 mole of Na2CO3.10H2O. Solution
Moles of oxygen atoms in 1 mole of Na2CO3.10H2O = 3+10 = 13
Moles of oxygen atoms in 0.2 mole of Na2CO3.10H2O = $0.2 \times 13= 2.6$
Therefore, Number of oxygen atoms = $2.6 \times 6.022 \times 10^{23}=1.565 \times 10^{24}$.
Question 32
A polystyrene, having the formula $Br_3C_6H_3(C_3H_8)_n$, was prepared by heating styrene with tribromobenzoyl peroxide in the absence of air. If it was found to contain 10.46% bromine by weight, find the value of n. Solution
Let the mass of polystyrene be 100g,
Therefore, No. of moles of Br in 100g of polystyrene = 10.46/79.9 = 0.1309
From the formula $Br_3C_6H_3(C_3H_8)_n$, we have,
$\text{No. of moles of Br} = 3 \times \text {moles of} \; Br_3C_6H_3(C_3H_8)_n$
$0.1309 = 3 \times \frac {mass}{\text{molecular mass}} = 3 \times \frac {100}{314.5+44n}$
n = 44.9
n=45(approx)
Question 33
It has been estimated that 93% of all atoms in the entire universe are hydrogen and that the vast majority of those remaining are helium. Based on only these two elements, estimate the mass percentage composition of the universe. Solution
It is given that out of 100 atoms, 93 atoms are Hydrogen and 7 atoms are Helium.
Mass of Hydrogen atoms = $93 \times 1 = 93$
Mass of Helium atoms = $7 \times 4 =28$
Therefore, Mass percentage of Hydrogen = $ \frac {93}{93+28} \times 100 =76.86$%
Mass percentage of Helium = $\frac {28}{93+28} \times 100 =23.14$%.
Question 34
The molecular weight of haemoglobin is about 65,000 g/mol. Haemoglobin contains 0.35% Fe by mass. How many iron atoms are there in a haemoglobin molecule? Solution
Mass of Fe in haemoglobin
= 0.35% of 65000
$= \frac {0.35 \times 65000}{100}$
= 227.5
Therefore, No. of Fe atoms in a haemoglobin molecule = $ \frac {227.5}{56} =4$
Question 35
Calculate the Number of atoms of oxygen present in 88g of $CO_2$. What would be the weight of the Carbon monoxide having the same number of oxygen atoms? Solution
Moles of $CO_2$ in 88 g = 88/44 = 2
Number of Oxygen atom=$ 2 \times 6.022 \times 10^{23} \times 2 = 20.092 \times 10^{23}$
Now
Number of molecules of CO having $20.092 \times 10^{23}$ of oxygen atom will be $20.092 \times 10^{23}$
Hence of weight of CO = $ \frac { 28 \times 20.092 \times 10^{23}}{6.022 \times 10^{23}}| =112$g
Question 36
At room temperature, the density of water is 1.0 g/mL and the density of ethanol is 0.789 g/mL. What volume of ethanol contains the same number of molecules as are present in 175 mL of water? Solution
Let the volume of ethanol containing the same number of molecules as are present in 175mL of water be V mL.
Moles of $C_2H_5OH$ in V mL = Moles of $H_2O$ in 175 mL
Mass of $C_2H_5OH$ /mol. mass of $C_2H_5OH$ = Mass of $H_2O$ / mol. mass of $H_2O$
Now as $Density= \frac {Mass}{volume}$ or $Mass = Density \times Volume$
Therefore
$ \frac {0.789 \times V}{46} = \frac {1 \times 175}{18}$
V =566.82 mL.
Question 37
Chlorophyll the green colouring matter of plants responsible for photosynthesis contains 2.68% of Magnesium by weight.Calculate the number of magnesium atoms in 4 g of
Chlorophyll Solution
Mass % of Mg = 2.68%
Now 100 g of Chlorophyll contains 2.68 g of Mg
Then 4 g of Chlorophyll contains = $(\frac {2.68}{100}) \times 4 = .1072$ g
Now
Number of atoms of Mg =$ \frac {Mass}{\text {Molar mass}} \times 6.022 \times 10^{23} = 1.34 \times 10^{21}$ atoms
Question 38
Calculate approximately the diameter of an atom of mercury, assuming that each atom is occupying a cube of edge length equal to the diameter of the mercury atom. The density of mercury is 13.6 g/cc. Solution
Suppose the side of cube = x cm = diameter of mercury atom
Therefore, Volume of 1 Hg atom = $x^3$ and
Mass of 1 Hg atom
$= density \times volume$
$= 13.6 \times x^3$
Mass of 1 Hg atom
= Atomic mass/ Avogadro constant
$= \frac {200}{6.022 \times 10^{23}}$
$13.6 \times x^3= \frac {200}{6.022 \times 10^{23}}$
$x = (2.44 \times 10^{-23})^{1/3}=2.9 \times 10^{-8}$ cm.
Question 39
How many years it would take to spend Avogadro number of rupees at the rate of Rs 10 laks per sec? Solution
Rupees= $6.022 \times 10^{23}$
Rupees spend rate = 106 /sec
Rupees spent per year =$10^6 \times 3600 \times 24 \times 365$
Number of Years required to fully spend the money
$=\frac {6.022 \times 10^{23}}{ 10^6 \times 3600 \times 24 \times 365}$
$=1.90988 \times 10^{10}$ years
Question 40
Find the total number of neutrons present in 7 mg of 14C atoms. Solution
1 mole of 14C=14 g
No. moles of carbon atoms = 7 mg/14 g = $5 \times 10^{-4}$ Moles
Number of atoms of carbon atoms = $6.022 \times 10^{23} \times 5 \times 10^{-4}= 3.01 \times 10^{20}$
A C-14 atom contains 8 neutrons
Therefore, The total number of neutrons = $8 \times 3.01 \times 10^{20} =2.408 \times 10^{21}$.
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