TEXT BOOK EXERCISE Q1. Select the most suitable answer ... · PDF file1st year chemistry notes Chapter No. 1 BASIC CONCEPTS TEXT BOOK EXERCISE Q1. Select the most suitable answer from

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Chapter No. 1

BASIC CONCEPTS

TEXT BOOK EXERCISE

Q1. Select the most suitable answer from the given ones in each question. (i) The mass of one mole of electrons is

(a) Properties which depend upon mass

(b) Arrangement of electrons in orbital

(c) Chemical properties

(d) The extent to which they may be affected in electromagnetic field

(ii) Which of the following statements is not true? (a) isotopes with even atomic masses are comparatively abundant

(b) isotopes with odd atomic masses and even atomic number are

comparatively abundant

(c) atomic masses are average masses of isotopes.

(d) Atomic masses are average masses of isotopes proportional to their

relative abundance

(iii) Many elements have fractional atomic masses, this is because (a) The mass of the atom is itself fractional

(b) Atomic masses are average masses of isobars

(c) Atomic masses are average masses of isotopes.

(d) Atomic masses are average masses of isotopes proportional to their

relative abundance

(iv) The mass of one mole of electrons is (a) 008mg (b) 0.55mg (c) 0.184mg (d) 1.673mg

(v) 27g of Al will react completely with how much mass of O2 to produce

Al2O3 (a) 8g of oxygen (b) 16g of oxygen

(c) 32g of oxygen (d) 24g of oxygen

(vi) The number of moles of CO2 which contain 8.0 g of oxygen. (a) 0.25 (b) 0.50 (c) 1.0 (d) 1.50

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(vii) The largest number of molecules are present in (a) 3.6g of H2 O (b) 4.8g of C2H5 OH

(c) 2.8 g of CO (d) 5.4g of N2O5

(viii) One mole of SO2 contains

(a) 6.02x1023

atoms of oxygen

(b) 18.1x1023

Molecules of SO2

(c) 6.02x1023 atoms of sulphur

(d) 4 gram atoms of SO2

(ix) The volume occupied by 1.4 g of N2at STP is

(a) 2.24 dm3

(b) 22.4dm3

(c) 1.12 dm3 (d) 112 cm

3

(x) A limiting reactant is the one which (a) is taken in lesser quantity in grams as compared to other reactants

(b) is taken in lesser quantity in volume as compared to the other reactants

(c) give the maximum amount of the product which is required

(d) give the minimum amount of the product under consideration

Ans: (i)a (ii)d (iii)d (iv)b (v)d (vi)a (vii)a (viii)c

(ix)c (x)d

Q2: Fill in the blanks: (i) The unit of relative atomic mass is-----------

(ii) The exact masses of isotopes can be determined by ------------

spectrograph.

(iii) The phenomenon of isotopes was first discovered by --------------

(iv) Empirical formula can be determined by combustion analysis for those

compounds which have-----------and -----------in them.

(v) A limiting reagent is that which controls the quantities of -------------

(vi) I mole of glucose has-----------atoms of carbon ---------------of oxygen

and ----------of hydrogen.

(vii) 4g of CH4 at OoC and I atm pressure has ---------molecules of CH4.

(viii) Stoichiometry calculations can by performed only when -------------law is

obeyed.

Ans: (i) amu (ii) mass (iii) Soddy (iv) carbon, hydrogen

(v) Products (vi) 6NA,6NA,12NA

(vii) 1.505x1023

(viii) conservation and multiple

proportion

Q3: Indicate true or false as the case may be:

(i) Neon has three isotopes and the fourth one with atomic mass 20.18 amu.

(ii) Empirical formula gives the information about he total number of atoms

present in the molecule

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(iii) During combustion analysis Mg(CIO4)2 is employed to absorb water

vapors.

(iv) Molecular formula is the integral multiple of empirical formula and the

integral multiple can never be unity.

(v) The number of atoms in 1.79 g of gold and 0.023g of sodium are equal.

(vi) The number of electrons in the molecules of CO an dN2 are 14 each, so

1 mg go each gas will have same number of electrons.

(vii) Avogadro’s hypothesis is applicable to all types of gases, i.e., ideal and

non-ideal .

(viii) Actual yield of a chemical reaction may by greater than the theoretical

yield.

Ans. (i) False (ii) False (iii) True (iv) False

(v) False (vi) True (vii) False (viii) False

Q4: What are ions? Under What condition are they produced?

Ans: Ions can be produced by the following processes:

(i) By dissolving ionic compounds in water

(ii) By X-rays

(iii) In mass spectrometry

(iv) By removing or adding electron in atom

Q4:

(a) What are isotopes? How do you deduce the fractional atomic masses of

elements form the relative isotopes abundance? Give two examples in

support of your answer. ( See detail in Sublime subjective)

(b) How does a mass spectrograph show the relative aboundace of isotopes of

an element? . ( See detail in Sublime subjective)

(c) What is the justification of two strong peaks in the mass spectrum for

bromine; while for iodine only one peak at 127 amu , is indicated?

Ans The two strong peak in the mass spectrum for bromine represent two

different isotopes of bromine having nearly equal natural abundances. Only

one peak at 127 amu in the mass spectrum for iodine indicates that it has

only one isotope of atomic mass 127 amu.

Remember that! Height of the peaks ∞ Relative abundance of isotopes

No. of peaks = No. of isotopes

Q5: Silver has atomic number 47 and has 16 known isotopes but two occur

naturally I,e, Ag _____107 . and Ag _____109 . Given the following mass

spectrometric data, calculated the average atomic mass of silver,

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Isotopes mass (amu) percentage abundance 107

Ag 106.90509 51.84 109

Ag 108.90476 48.16

Solution: The mass contribution for silver are:

Isotopes Fractional abundance isotopic mass mass contribution

107Ag 107 0.5184x107=55.4688

109Ag 107 0.4816x109=52.4944

Fractional atomic mass of silver =107.9632

Hence the fractional atomic mass of silver is =107.9632 Ans.

Q6: Boron with atomic number 5 has two naturally occurring isotopes.

Calculate the percentage abundance of 10

B and 11

B from the following

information.

Average atomic mass of boron =10.81 amu

Isotopic mass of 10

B =10.0129 amu

Isotopic mass of 11

B =11.0093

Solution: Let, the fractional abundance of 10

B =x

The fractional abundance of 11

B =1-x

Remember that the sum of the fractional abundances of isotopes must be

equal to one, now, The equation to determine the atomic mass of element is

(fractional abundance) (isotopic mass) (fractional abundance of 10

B)(isotopic

mass of 10

B )+(fractional abundance of 11

B) (isotopic mass of 11

B)

=Average atomic mass of Boron

(x)(10.0129)+(1-x)(11.0093) =10.81

10.0129x+11.00093x =10.81

10.0129x-11.00093x =10.81-11.0093

-0.9964x =-0.1993

x =

Fractional abundance of 10

B =0.2000

Fractional abundance of 11

B =(1-0.2000)=0.8000

By percentage the fractional abundance of isotope is

%of 10

B =0.2000x100 =20% Answer

% of 11

B =0.8000x100 =80%Answer

Q7: Define the following terms and give three examples of each.

(i) Gram atom (ii) Gram molecular mass

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(iii) Gram molecular mass (iv) Gram ion

(v) molar volume (vi) Avogadro’s number

(vii) Stoichiometry (viii) Percentage yield

Q8: Justify the following statements:

(a) 23 g of sodium and 238g of uranium have equal number of atoms in the (b)

Mg atom is twice heavier than that of carbon

(c) 180g of glucose and 342 g of sucrose have the same number of molecules

but different number of atoms present in them.

(d) 4.9g of H2 SO4 when completely ionized in water , have equal number of

positive and negative charges but the number of positively charged ions are

twice the number of negatively charged ions.

(e) One mg of K2 Cr O4 has thrice the number of ions than the number of

formula units when ionized in water.

(f) Two grams of H2 , 16 g of ch4 and 44g of CO2 occupy separately the

volumes of 22.414 dm3

, although the sizes and masses of molecules of three

gases are very different from each other.

Solution:

(a) 23g of Na =1 mole of Na =6.02x1023

atoms of Na

238g of U =1 mole of U =6.02x1023

atoms of U.

Since equal number of gram atoms(moles) of different elements contain

equal number of atoms. Hence, 1 mole (23g ) of sodium and 1 mole (238)g of

uranium contain equal number of atoms , i , e ,6.02x1023

atoms.

(b) Since the atomic mass of Mg (24) is twice the atomic mass of carbon (12)

therefore, Mg atom is twice heavier than that of carbon. Or

Mass of 1 atom of Mg=

Mass of 1 atom of C =

Since the mass of one atom of Mg is twice the mass of one atom of C, therefore,

Mg atom is twice heavier than that of carbon.

(c) 180 g of glucose = 1 mole of glucose =6.02x1023

molecules of glucose 342

g of sucrose=1mole of sucrose =6.02x1023

molecules of sucrose

Since one mole of different compounds has the same number of molecules.

Therefore 1 mole (180g) of glucose and I mole (342g) of sucrose contain the

same number (6.02x1023

)of molecules. Because one molecule of glucose, C6H12O6

contains 45 atoms whereas one molecules of glucose, C12 H22 O11 contains 24

atoms. Therefore, 6.02x1023

molecules of glucose contain different atoms as

compound to6.02x1023 molecules of sucrose. Hence, 180 g of glucose and 342g og

sucrose have the same number of molecules but different number of atoms present

in them.

(d) H2 SO4 2H+

+ SO

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When one molecules of H2 SO4 completely ionizes in water it produces two

H+

ion and one SO ion,. Hydrogen ion carries a unit positive charge whereas SO

ion carries a double negative charge. To keep the neutrality, the number of

hydrogen are twice than the number of soleplate ions. Similarly the ions produced

by complete ionization of 4.8g of H2 SO4 in water will have equal number of

positive and negative but the number of positively charged ions are twice the

number of negatively charged ions.

(e) H2 SO4 2H+

+ SO

K2 Cr O4 when ionizes in water produces two k+

ions one C O ion. Thus

each formula unit of K2 Cr O4produces three ions in solution .Hence one mg of K2

Cr O4 has thrice the number of ion than the number of formula units ionized in

water.

(f) 2g of H2 =1 mole of H2 =6.02x1023 molecules of H2 at STP =22.414dm3 16g

of CH4 =1mole of CH4 =6.02x1023 molecules of CH4 at STP =22.414dm3

144g of

CO2 =1mole of CO2 =6.02x1023 molecules of CO2 at STP =22.144dm3

Although H2 , CH4 and CO2 have different masses but they have the same

number of moles and molecules . Hence the same number of moles or the same

number of molecules of different gases occupy the same volume at STP. Hence 2 g

of H2 ,16g of CH4 and 44 g of CO2 occupy the same volume 22.414 dm3

at STP.

The masses and the sizes of the molecules do not affect the volumes.

Q10: Calculate each of the following quantities

(a) Mass in grams of 2.74 moles of KMnO4 .

(b) Moles of O atoms in 9.0g of Mg (NO3)2 .

(c) Number of O atoms in 10.037g of Cu SO4 .5H2 O.

(d) Mass in kilograms of 2.6x 1020

molecules of SO2 .

(e) Moles of C1 atoms in 0.822g C2H4C12 .

(f) Mass in grams of 5.136 moles of silver carbonate .

(g) Mass in grams of 2.78x1021

molecules of CrO2 C12 .

(h) Number of moles and formula units in 100g of KC1O3 .

(i) Number of K+

ions C1O ions, C1 atoms, and O atoms in (h)

Solution: (a) No of moles of KMnO4 =2.74moles

formula mass of KMnO4 =39+55+64=158g mol -1

Mass of KMnO4 =?

Formula used:

Mass of KMnO4 = no of mole of KMnO4 x formula mass of KMnO4

=2.74 mol x 158 g mol-1

=432.92g Answer

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(b) Mass of Mg (NO3)2 =9g

Formula mass of Mg (NO3)2 =24+28+96=148g mol -1

No of moles of O atoms =?

Formula used:

No of mole of Mg (NO3)2 =

Now, I mole of Mg (NO3)2 contains = 6moles of O atoms

0..06 moles of Mg (NO3)2contains =6x0.6

=0.36 moles of O atoms

Alternatively,

148g of Mg (NO3)2 contains =6moles of O atoms

g of Mg (NO3)2contains =

=0.36 mole Answer

(c) Mass of CuSO4. 5H2O=10.037g

Formula mass of CuSO4. 5H2O=63.54+32+64+90

=249.546g mol -1

No of moles of CuSO4. 5H2O =?

No of moles of CuSO4. 5H2O =

=

Now, 1 mole of CuSO4 .5H2O contains 9moles of O atoms

0.04 mole of CuSO4 .5H2O contains=9x0.04

=0.36 moles of O atoms

Now, I mole of O atoms contains =6.02x1023

O atoms

0.36 mole of O atoms contains =6.02x1023

x0.36 oxygen

atoms

=2.17x1023

oxygen atoms

=2.17x1023

atoms Answer

(d) No of molecules of SO2 . =2.6x1020

molecules

Molecular mass of SO2 . =32+32=64 g mol-1

Now, Avogadro’s number , NA =6.02x1023

molecules of SO2

Mass of SO2 molecules

=27.64x10-3 g

=

=27.64x10-6

kg

=2.764x10-3

kg Answer

(e) Mass of C2 H4C1 = 0.822g

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Molecular mass of C2 H4C1 =24+4+71=99 g mol-1

No of moles of C2 H4C1 =

Now, 1 mole of C2 H4C1 contains =2moles of C1 atoms

8.3x10-3mole of C2 H4C1 contains =2x8.3x10-3

mole of atom

=16.6x10-3

=0.0166mole of C1 atom

=0.017 mole Answer

(f) No of mole of Ag2 CO3 =5.136moles

Formula mass of Ag2 CO3 =215.736+12+48=275.736 g mol-1

Mass of Ag2 CO3=No of moles of Ag2 CO3xformula mass of Ag2 CO3

=5.136molx275.736 g mol-1

=416.18g

=1416.2 g Answer

(g) Molecular mass of CrO2C12 =52+32+71=155g mol-1

NA =6.02x1023

molecules mol-1

Molecules of CrO2C12==2.78x1021

molecules

Now, mass of CrO2C12

= =

=71.578x10-2

g

=0.71578

=0.716 g Answer

(h) Mass of KCIO3 =100g

Formula mass of KCIO3 =39x35.5+48=122g mol-1

No of moles of KCIO3 =?

No of moles of KCIO3 =

= =0.816mole Answer

No of formula units No of moles x Avogadro,s No

=0.816mole x 6.02x1023

formula units

=4.91x1023

formula units

(i) No of K+

ions =4.91x1023

Answer

No of CIO ions =4.91x1023

Answer

No of CIO ions =4.91x1023

Answer

No of O atoms = 4.91x1023

x3

=14.73x1023

=1.473x1024

Answer

Q 11 Aspartame he artificial sweetener, has a molecular formula of C14 H18

N2O5 .

(a) What is the mass of one mole of aspartame?

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(b) How many moles are present in 52g of aspartame?

(c) What is the mass in grams of 10.122 moles of aspartame?

(d) How many hydrogen atoms are present in 2.34g of aspartame?

(a) Molecular mass of aspartame =168+18+28+80=295g mol-1

Mass of 1 mole of aspartame =294g mol-1

Answer

(b) Mass of aspartame =52g

Molecular mass of aspartame =294g mol-1

No of moles of aspartame =

=

=0.1768 mol

=0.177 mol Answer

(c) No moles of aspartame = 10.122 moles

Molecular mass of aspartame =294g mol-1

Mass of aspartame =No of moles x Molar mass

=10.122mol x 294g mol-1

=2975.87 g Answer

(d) Mass of aspartame =243g

Molar mass of aspartame =294g mol -1

No of molecules of aspartame=?

No of molecules of aspartame= xNA

=

=

=4.98x1021

molecules.

Now,1 molecule of aspartame contains =18 H atoms

4.98x 1022 molecules =18x4.98x1021

H atoms

=89.64x1021

H atoms

=8.964x1022 H atoms Answer

Q 12: A sample of 0.600 mole of a metal M reacts completely with excess of

fluorine to from 46.8g MF2 .

(a) How many moles of F are present in the sample of MF2 that forms.

(b) which elements is represented by the symbol M ?

Solution: (a) Formula of compound =MF2

No of moles of M =0.6 mol

Mass of MF2 =46.8g

The molar of M:F in the compounds;

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No of moles of F =0.6x2=1.2mol Answer

Mass of F =No of moles of Fx At . mass of F

=1.2x19=22.8g

Mass of compound =46.8g

Mass of metal, M =46.8-22.8

=24

At mass of M =

=

(b) The atomic mass of the elements, M =40

The metal is calcium, Ca Answer

Q 13 : In each pair , choose the larger of the indicated quantity ,or state if the

samples are equal.

(a) Individual particles: 0.4 mole of oxygen molecules or0.4mole of

oxygen atom.

(b) Mass: 0.4 mole of ozone molecules or0.4mole of oxygen atoms

(c) Mass: 0.6 mole of C2 H4 or 0.6mole of 12

(d) Individual particles: 4.0g N2O4 or 3.3g SO2

(e) Total ions: 2.3 moles of NaC1O3 or 2.0mole of MgC12

(f) Molecules: 11.0g of H2Oor 11.0g H2O2

(g) Na+

ion: 0.500 moles of NaBr or 0.0145kg NaC1

(h) Mass: 6.02x1023

atoms of 235

U or 6.02x1023 atoms of 238

U

Ans: (a) Number of molecules =moles x NA

Number of O2 molecules =0.4x6.02x1023

=2.408x1023

molecules

No of O atoms=0.4x6.02x1023

=2.108x1023

atoms

There are equal number of individual particles in 0.4 mole of oxygen

molecules and 0.4 mole of oxygen atom. In general, equal number of moles of

different substances contains equal number of particles.

Both are equal Answer (b) Mass of substance = moles x molar mass

Mass of oxygen atoms =0.4x16=64g

Mass of ozone, O3 molecules =0.4x48=19.2g

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0.4 moles of ozone molecules have larger mass than 0.4mole of oxygen atoms.

Ozone Answer (c) Mass of C2H4 =0.6x28=1.68g

Mass of 12 =0.6x127=254g

0.6mole of 12 have larger mass than 0.6 mole of C2H4

12 Answers (d) No of molecules =

No of molecules in N2 O4 = x6.02x1023

=2.62 x1023

molecules

No of molecules in SO2 =x6.02x1023

=3.1x1022

molecules

3.3g of SO2 have larger number of individual particles than 4.0 g of N2 O4 .

SO2 Answer (e) No of formula units =Moles x NA

No of formula units of NaC1O3 =2.3x6.02x10

23=1.38x10

24 formula

units

No of ions in 1 formula units of NaC1O3=2

Total no of ions in MgC12 =2x1.38x1023

=2.76x1024

ions

No of formula units of MgC12 =2.0x6.02x1023

x3=3.6x1024 ions

No .of ions in one formula unit of MgC12 =3

Total no of ions in MgC12 =1.20x1024

x3=3.6x1024

ions

2.0moles of MgC12 contain lager number of total ions than 2.3 moles of NaC1o3-

MgC1 Answer

(f) No of molecules = NA

No of molecules in H2 O2 = x6.02x1023

=3.68x1023

molecules

No of molecules in H2 O2 = x6.02x1023

=1.95x1023

molecules

11.0g of H2 O2contains larger number of molecules than 11.0g of H2 O2

H2 O2Answer (g) No of formula units =moles xNA

No of formula units NaBr =0.5x6.02x1023

=3.01x1023

formula units

One formula units o NaBr contain Na+

ions =1

3.01 x1023

formula unit of NaBr contains Na +ions =3.01x10

23 Na

+ ions

No of formula units of NaC1 = x6.02x1023

=1.49x1023

formula

units

One formula unit of NaC1 contains Na+

ions =1

1.49x1023

formula units of NaC1 contains =1.49x1023

Na+

ions

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0.500 moles of NaBr contains lager number of Na+

ions than 0.0145kg ofNaC1.

NaBr Answer (h) Mass of atoms of an element =

Mass of 235

Uatoms =x6.02x1023

=235g

Mass of 238

U atoms =x6.02x1023

=238g 238

U Answer Q 13:

(a) Calculate the percentage of nitrogen in the four important fertilizer i.e.,

(i)NH3 (ii)NH2CONH2(Urea) (iii)(NH4)2SO4 (iv)NH4 NO3

(b) Calculate the percentage of nitrogen and phosphorus in each of the

following:

(i) NH4H2PO4 (ii) (NH4)) PO4 (iii) (NH4)4 PO4

Solution: (a) Mol-mass of NH3 =14+4=17g

Mass of N =14g

% of N =x100

=82.35% Answer

(b) Mol-mass of NH2 CONH2 =28+4+12+16=60g

Mass of N =28g

%of N =x100

=46.35% Answer (c) Mol-mass of (NH2 )2 SO4 =28+8+32+64=132g

Mass of N =28g

% of N =x100

=21.21% Answer (d) Mol-mass of (NH2 )2 SO4 =28+4+48=80g

Mass of N =28g

%of N =x100

=35% Answer

(I) Mol-mass of (NH2 )2 SO4 =14+6+31+64=115g

Mass of N =14g

Mass of P =31g

%of N =x100=12.17% Answer

%of P ==26.96% Answer

(II) Mol-mass of ((NH2 )2 SO4 =28+9+31+64=132g

Mass of N =28g

Mass of P =

%of N = =21.21% Answer

%of P = =23.48% Answer

(III) Mol-mass of (NH2 )2 SO4 =42+12+31+64=149g

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Mass of N =42g

Mass of P =31g

%of N =

%of P =

Q 14: Glucose C6 H12 O6 is the most important nutrient in the cell for generating

chemical potential energy. Calculate the mass% of each element in glucose and

determine the number of C,H and O atoms in 10.5g go the sample.

Solution: Mol-mass of glucose C6 H12 O6 =72+12+96=180g

Mass of C =72

Mass of H =12

Mass of O =96

% of C = =40% Answer

% of H = =6.66% Answer

% of O = =53.33% Answer

Mass of C6 H12 O6 =10.5g

Mol-mass of C6 H12 O6 =180g

Mol-mass of =180g mol-1

No of moles of C6 H12 O6 =

No of molecules of glucose =No of moles x NA

=0.058 molx 6.02x1023

molecules mol-1

=0.35x1023

molecules

=3.5x1022

molecules

Now, 1 molecule of glucose contains =6C-atoms

3.4x1022

molecules of glucose contains =6x3.5x1022

C-atoms

=21x1022

=2.1x1023

C atoms Answer

1 molecules of glucose contains =12H-atoms

3.5x1022

molecules glucose contains =12x3.5x1022

=4.2x1023

H- atoms Answer

1 molecule of glucose contains =6 O –atoms

3.5 x 1022

molecules of glucose contains =6x3.5x1022

=2.1x1023

O-atoms Answer

Q 16: Ethylene glycol is used as automobile antifreeze .It has 38.7% carbon,

9.7% hydrogen and 51.6% oxygen. Its molar mass is 62.1 grams mol-1

.Determine

its molecular formula.

Solution: % of C=38.37 g % of H =9.7g % of O=51.6g

At. Mass of C=12g mol-1

At. Mass of H=1.008g mol-1

At. Mass of O =16g

mol-1

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No of moles of C =

No of moles of H =

No of moles of O =

Atomic ratio is obtained by dividing the moles with 3.23, which is the smallest

ratio.

C :H :O

1 :3 :1

Empirical formula =CH3 O

Empirical formula mass =31

n=

Molecular formula =2x CH3 O

=C2 H6 O2 Answer

Q 16: Serotonin (Molecular mass= 176g mol-1

) is a compound that conducts

nerve impulses in brain and muscles. It contains 68.2 % C, 6.86% H, and 9.08%

O. What is its molecular formula?

Solution:

No of moles of C =

No of moles of H =

No of moles of N =

No of moles of O =

C : H : N : O

Atomic

Ratio

: : :

10 : 12 : 2 : 1

Empirical formula =C10 H12 N2 O

Empirical formula mass =120+12+28+16=176g mol-1

Molecular mass =176g mol-1

n=

Q17: An unknown metal M reacts with S to from a compound with a formula

M2S3 .If 3.12 g of M reacts with exactly 2.88 g of sulphur, what are the names of

metal M and the compound M2 S3 .

Solution: Formula of compound = M2 S3

Mass of M =3.12g

Mass of S =2.88g

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Atomic mass of S =32g mol-1

No of moles of S =

No of moles of S =

The molar ratio of M: S in the compound is :

No of moles of M =

=0.06 mole

Now, No of moles of M =

At. Mass M =

The mass of M used in the formation of M2S3 is 3.12g. The product M2S3 therefore

also contains 3.12g of M, because mass is conserved. The amount of M before and

after reaction must be the same. Since we know both the number of moles of M

and the mass of M , we can cal calculate the atomic mass of M as follows:

At. Mass of M =

=52

Atomic number, Z =52

Q19: The octane present in gasoline burns according to the following equation.

2C8 H18 (i) + 2502(g) 16CO 2(g) + 18H2O (i)

(a) How many moles of O2 are needed to react fully with 4 moles of

actane?

(b) How many moles of CO2 can be produced from one mole of actane?

(c) How many moles of water are produced by the combustion of 6 moles

of octane?

(d) If this reaction is to be used to synthesize 8 moles of CO2 how many

grams of oxygen are needed? How many grams of octane will be used?

Solution: 4 moles

2C8 H18 (i) + 2502(g) 16CO 2(g) + 18H2O (i)

(a) 2 moles 25 moles

2 moles of C8 H18 =25 moles of O2

4 moles of C8 H18 =

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=50moles of O2 Answer

(b) 1 moles

2C8 H18 (i) + 2502(g) 16CO 2(g) + 18H2O (i) 2 moles

Now, 2 moles of C8 H18 =16 moles of CO2

1 mole of C8 H18 =

=8 moles of CO2 Answer

(c) 6 moles

2C8 H18 (i) + 2502(g) 16CO 2(g) + 18H2O (i) 2 moles

Now, 2 moles of C8 H18 =18 moles of H2 O(i)

6 moles of C8 H18 =

=54 moles of H2 O

(d) 6 moles

2C8 H18 (i) + 2502(g) 16CO 2(g) + 18H2O (i) 2 moles 1800moles

Now, 16 moles of CO2 =25 moles of O2

8 moles of CO2 =

=12.5 moles of CO2

Mol-mass of O2 =32g mol-1

=12.5 molx 32g mol-1

=400g of O2

Now, 16moles of CO2 =2moles of C8 H18

8 moles of CO2 =

=1 mole of C8 H18

Mol-mass of C8 H18 =96+18=114g mol-1

Mass of C8 H18 =No of moles of C8 H18xMol.mass ofC8

H18 =1 molx 114 g mol-1

114g Answer

Q19: Calculate the number of grams of A12 S3 which can be prepared by the

reaction of 20 g of A1 and 30 g of sulphur. How much the non-limiting reaction is

in excess?

Solution: Mass of A1 =20g

Molar mass of A1 =27g mol-1

No of moles of A1 =

Mass of S = 30g

Molar mass of S =32g mol-1

No of moles of S =

0.74 mole 0.94 mole

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2A1 + 3S A12 S3

2 mole

3 mole

1 mole

Now, 2 moles of A1 =1 mole of A12 S3

0.74 moles of A1 =

=0.37 mole of A12 S3

Now, 3 moles of S =1 moles of A12 S3

0.94 moles of S =

=0.313 mole of A12 S3

Since S give the least number of moles of A12 S3 therefore, it is the limiting

reactant.

No of moles of A12 S3 =0.313 mole

Molar mass of A12 S3 =150g mol-1

Mass of A12 S3=No of moles of A12 S3xMolar mass of A12 S3

=0.313molx 150 g mol-1

=46.95 g of A12 S3 Answer

The non-limiting reactant is A1 which is in excess. Now mass of A1 required

reacting completely with 0.94 moles of S can be calculated as:

0.94 mole

2A1 + 3S A12 S3

2 mole

3 mole

Now, 3 moles of S =2 moles of A1

0.94 moles of S =

=

Mass of A1 =No of moles of A1 x molar mass of A1

=0.63x 27

=17g of A1

Mass of A1available =20g

Mass of A1 which reacts completely =17g with available S

Excess of A1 =20-17=3g

Q20: A mixture of two liquids, hydrazine N2H4 and N2 O4 are used as a fuel in

rockets. They produce N2 and water vapors. How many grams of N2 gas will be

formed by reacting 100g of N2 O4 and 200g g of N2 O4.

2N2H4 + N2O2 3N2 +4 H2O

Solution: Mass of2N2H4 =100g

Mass of N2O2 =200g

Molar mass of 2N2H4 =28+4=32g mol-1

Molar mass of N2O2 =28+64=92g mol-1

No of moles of N2H4 =

No of moles of N2O2 =

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3.125moles 2.174 moles

2N2H4 + N2O2 3N2 +4 H2O 2 moles 1mole 3moles

Now , 2moles of N2H4 =3moles of N2

3.125moles of N2H4 =

=4.69 mole of N2

Now , 1 mole of N2O2 =3moles of N2

2.174 moles of N2O4 =

=6.52 mole of N2O2

Since N2H4gives the least number of moles of N2, hence it is the limiting reactant.

Amount of N2 produced =4.69 moles

Molar mass of N2

=28g mol-1

Mass of N2 =4.69g molx 28g mol-1

=131032 g Answer

Q21: Silicon carbide (SiC) is an important ceramic material . It is produced by

allowing sand (SiO2 )to react with carbon at high temperature.

SiO2 + 3C SiC + 2CO

When 100kg sand isn reacted with excess of carbon, 51.4 kg of Sic is produced.

Solution: Mass of SiO2 =100 kg=100000g

Mass of SiC produced =5.14 kg =51400g

100000g

SiO2 + 3C SiC + 2CO

60g

40g

Now, 60g of SiO2 =40g of SiC

100000g of SiO2 =

=66666.67 g

Actual yield of Sic =51400 g

Theoretical yield of SiC =66666.67g

% yield =

=

=77.1%

Q22: (a) What is Stoichiometry? Give its assumptions? Mention two

important law, which help to perform the Stoichiometry calculations.

(b) What is a limiting reactant? How does it control the quantity of

the product formed? Explain with three examples

Q 23: (a) Define yield. How do we calculate the percentage yield of a

chemical reaction?

(b) What are the factors which are mostly responsible for the low yield

of the products in chemical reactions.

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Q24: Explain the following with reasons.

(j) Law of conservation of mass has to be obeyed during Stoichiometric

calculations.

(ii) Many chemical reactions taking place in our surrounding involves the

limit reactants.

(iii) No individual neon atom in the sample of the element has a mass of

20.18amu.

(iv) One mole of H2 SO4 should completely react with two moles of NaOH.

How does Avogadro, s number help to explain it.

(v) One mole H2 O has two moles of bonds , three moles of atoms , ten

moles of electrons and twenty eight moles of the total fundamental particles

present in it.

(vi) N2 and CO have the same number of electrons, protons and neutrons.

Ans. (i) According to law of conservation of mass, the amount of each

element is conserved in a chemical reaction. Chemical equations are written

and balanced on the basis of law of conversation of mass. Stoichiometry

calculations are related with the amounts of reactants and products in a

balanced chemical equation. Hence, law of conservation of mass has to be

obeyed during stoichiometric calculations.

(ii) In our surrounding many chemical reactions are taking place which

involve oxygen. In these reactions oxygen in always in excess quantity while

other reactant are in lesser amount. Thus other reactants act as limiting

reactants.

(iii) Since the overall atomic mass of neon in the average of the determined

atomic masses of individual isotopes present in the sample of isotopic

mixture .Hence, no individual neon atom in the sample has a mass of

20.18amu.

(iv) H2 SO4 +2NaOH Na2 SO4 + 2H2 O 1 mole 2moles

2 moles of H+

ions 2 moles of OH ions

2x6.02x1023

H+

ions 2x6.02x1023

OH ions

Once mole of H2SO4 consists of 2 moles of H+

ions that contains twice the

Avogadro’s number of H+

ions. For complete neutralization it needs 2 moles

of one mole of H2 SO4 should completely react with two moles of NA OH.

(v) Since one molecule of H2O has two covalent bonds between H and O

atoms. Three atoms, ten electrons and twenty eight total fundamental

particles present in it. Hence, one mole of H2 O has two moles of bond, three

moles of atoms, ten moles of electrons and twenty eight moles of total

fundamental particle present in it.

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In N2 there are 2 N atoms which contain 14 electrons (2x7),14 protons (2x7) and 14

neutrons (2x7) . In CO, there are one carbon and one oxygen atoms. It contains 14

electrons (6carbon e +8 oxygen e), 14 protons (6 C proton +8 O proton ) and 14

neutrons (6 neutrons +8 O neutrons).Hence , N2 and CO have the same number of

electrons, protons and neutrons. Remember that electrons, protons and neutrons of

atoms remain conserved during the formation of molecules in a chemical reaction.

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