Mole Concepts Problems with Solutions

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

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 $NH_4Cl$
$= \text {No. of moles} \times \text {molar mass}$
$= 1 \times 53.5g =53.5g$.

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

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
$= 16 x 6.022 \times 10^{23}=96.352 \times 10^{23}$

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.

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.

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.

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.

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}}$
$= \frac {40 \times 1000}{40} =1000$

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

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.

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

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 = $2 \times \text{No. of molecules} =2 \times 3.011 \times 10^{23}=6.022 \times 10^{23}$.

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

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.

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 = $4 \times 9.411 \times 10^{23}= 3.74 \times 10^{24}$

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.

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

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.

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

1 gm atom = atomic weight
So number of atoms will be Avogadro number = $6.022 \times 10^{23} $

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.

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

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

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
$n = \frac {pV}{RT}$
$n = \frac {(10^{-12} \times 1)}{(0.0821 \times 273)} = 0.44 \times 10^{-13}$
$\text {No. of molecules} = \text {no. of moles} \times \text {Avogadro constant}$
$= 0.44 \times 10^{-13} \times 6.022 \times 10^{23}$
$=2.65 \times 10^{10}$ .

Volume ratio
= Molar ratio (Avogadro’s principle – the molar ratios are also volume ratios for gases)
=1:1

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

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

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 = $2.6 \times 6.022 \times 10^{23}=1.565 \times 10^{24}$.

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)

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 = $\frac {28}{93+28} \times 100 =23.14$%.

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$

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

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.

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

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.

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

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 = $8 \times 3.01 \times 10^{20} =2.408 \times 10^{21}$.