Given below are the Mole Concepts Questions with Detailed Solutions

a. Concepts questions

b. Calculation problems

c. percentage composition

d. Mole fraction & molarity

a. Concepts questions

b. Calculation problems

c. percentage composition

d. Mole fraction & molarity

How many molecules of water are there in 54 g of $H_2O$ ?

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

Calculate the mass of $6.022 \times 10^{23}$ molecule of $NH_4Cl$ ?

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 $NH_4Cl$

$= \text {No. of moles} \times \text {molar mass}$

$= 1 \times 53.5g =53.5g$.

Calculate the mass of $12.044 \times 10^{23}$ Oxygen atoms.

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

How many atoms of hydrogen are there in 36 g of $NH_4$?

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

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

Calculate the number of hydrogen atoms in 1 mole of H

1 molecule of H_{2}= 2 hydrogen atoms

So, 1 mole of H_{2}= 2 mole hydrogen atoms

$= 2 \times 6.022 \times 10^{23}=12.044 \times 10^{23}$ hydrogen atoms.

Calculate the number of Cu atoms in 0.3175 g of Cu.

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.

Calculate the number of molecules in 22.4 litres of CH

1 mole of CH_{4}= 22.4 L (at NTP)

Therefore

22.4 L of CH_{4}gas_{}= 1 mole CH_{4}gas_{}= $6.022 \times 10^{23}$ CH_{4} gas molecules.

a. Find the volume of 1 g of H

b.Find the volume of 20g H

c.What is the volume occupied by $6.022 \times 10^{23}$ molecules of any gas at NTP?

a. Molar Mass of H_{2}gas = 2 gm

$\text {Number of moles} = \frac {Mass}{\text {Molar Mass}} = \frac {1}{2} = .5$

1 mole of H_{2}= 22.4 L (at NTP)

Therefore

.5 mole of H_{2}= .5 x 22.4 = 11.2 litre

b.

No. of moles of H_{2}= 20/2 =10

1 mole of H_{2}= 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.

An atom of some element Y weighs $6.644 \times 10^{-23}$ g. Calculate the number of gram-atoms in 40 kg of it.

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}}$

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

Find the number of moles and number of atoms of H and S in 10 mole of $H_2S$.

1 mole of H_{2}S contains 2 mole of H, 1 mole of S

Therefore

10 mole of H_{2}S 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

Calculate the number of atoms of each element in 245 g of $KClO_3$.

Molecular mass of KClO_{3}= $39 +35.5+3 \times 16 = 122.5$

No. of mole of KClO_{3}= 245 g/122.5g = 2 mole

2 mole of KClO_{3}contains

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.

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)

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

Calculate the number of atoms in 12.2 litres of the below gas at NTP

(i) mono-atomic

(ii) diatomic gas

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

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?

Mass of one mole of oxygen molecule (O_{2})

= molecular mass of oxygen molecule (O_{2}) in gram

= 32 g.

Mass of a single oxygen molecule

$= \frac {32}{1 \times 10^{24}}$

$=3.2 \times 10^{-23}$ g

Calculate the standard molar volume of oxygen gas. The density of $O_2$ gas at NTP is 1.429 g/L.

Standard molar volume is Volume of 1 mole of gas i.e 32 gm O_{2} gas

Now

$Density = \frac {Mass}{volume}$

$Volume= \frac {Mass}{Density}$

= 32/1.429

=22.39 Litres.

if 2 mol of Calcium Carbonate (Formula Weight =100) occupies a volume of 67.0 ml, Find the density ?

Mass of Calcium Carbonate = 2 * 100 = 200 g

Volume = 67.0 ml

$Density = \frac {Mass}{volume} = \frac {200}{67}= 2.99$ g/mL

Calculate how many methane molecules and how many carbon and hydrogen atoms are there in 25 g of Methane?

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

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.

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.

Calculate the number of moles of Ca, C and oxygen atoms and its mass in 200 g of $CaCO_3$.

Molecular mass of CaCO_{3}= 40+12+3 \times 16 = 100

No. mole of CaCO_{3}= $ \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

Calculate the total number of electrons present in 3.2 g of CH

Molecular mass of CH_{4}= $12+4 \times 1 = 16$

Moles of CH_{4}=$\frac {3.2}{16} =0.2$

No. of electron in 1 molecule of CH_{4}= 6+4 = 10 electrons

Total no. of electrons = $0.2 \times 6.022 \times 10^{23} \times 10 =12.044 \times 10^{23}$ electrons.

How much Calcium is in the amount of Ca(NO

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

State the number of atoms in 1 g atom of Aluminium?

1 gm atom = atomic weight

So number of atoms will be Avogadro number = $6.022 \times 10^{23} $

If the components of the air are N

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.

How many molecules are there in 1.624 gm Ferric chloride(FeCl

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

What is the mass of 1 Ammonia Molecule ?

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

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?

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

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.

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

$n = \frac {pV}{RT}$

$n = \frac {(10^{-12} \times 1)}{(0.0821 \times 273)} = 0.44 \times 10^{-13}$

$\text {No. of molecules} = \text {no. of moles} \times \text {Avogadro constant}$

$= 0.44 \times 10^{-13} \times 6.022 \times 10^{23}$

$=2.65 \times 10^{10}$ .

What is the ratio of the volumes occupied by 1 mole of O

Volume ratio

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

=1:1

Calculate the mass of .5 moles of CaCO

Molar mass (i.e., molecular mass in g) = $40+12+3 \times 16 = 100$g

Mass of .5 moles of CaCO_{3}= $.5 \times 100 = 50$ g

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?

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

Calculate the number of oxygen atoms in 0.2 mole of Na

Moles of oxygen atoms in 1 mole of Na_{2}CO_{3}.10H_{2}O = 3+10 = 13

Moles of oxygen atoms in 0.2 mole of Na_{2}CO_{3}.10H_{2}O = $0.2 \times 13= 2.6$

Therefore, Number of oxygen atoms = $2.6 \times 6.022 \times 10^{23}=1.565 \times 10^{24}$.

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.

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 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.

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

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?

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$

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?

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

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?

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.

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

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

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.

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.

How many years it would take to spend Avogadro number of rupees at the rate of Rs 10 laks per sec?

Rupees= $6.022 \times 10^{23}$

Rupees spend rate = 10^{6} /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

Find the total number of neutrons present in 7 mg of

1 mole of ^{14}C=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 = $8 \times 3.01 \times 10^{20} =2.408 \times 10^{21}$.

**Notes**- Some Basic Concepts of Chemistry
- Physical States of matter
- Chemical Classification Of matter
- The International System of Units (SI units)
- Significant Figures
- Laws of Chemical Combination
- Atomic Mass
- Molecular Mass
- Mole Concept (Avogadro Constant) And Molar mass
- Gram Atomic Mass
- Gram Molecular Mass
- Percentage composition
- StioChiometry
- Mole Fraction
- Molarity Definition
- Molality Definition
- Normality Formula

**Questions**