NCERT SOLUTION CLASS 11th CHEMISTRY UNIT 1 Some Basic Concepts of Chemistry

 NCERT SOLUTION CHEMISTRY CLASS 11 UNIT- 1 Some Basic concepts of Chemistry

. Topic Name

1. Importance of Chemistry

2. Nature of matter

3. Properties of Matter and their Measurement

4. Uncertainty in Measurement

5. Laws of Chemical combinations

6. Dalton s Atomic Theory

7. Atomic And Molecular Masses

8. Mole concept and Molar Masses

9. Percentage Composition

10. Stoichiometry and Stoichiometric Calculations


     NCERT TEXTBOOK EXERCISES QUESTION SOLVED

Q. Calculate the molecular mass of the following:

1. H2O 2. Co2 3. CH4

ANS. Molecular mass of H2O

   = 2(1.008u) + (160.00u) =18.016


2. Molecular mass of co2

 = 1 (12.01 u) +2(16.00u) =44.01u


3. Molecular mass of

= 1 (12.01u) +4(1.008u) = 16.042u


Q. Calculate the mass percent of different elements present in sodium sulphate (Na,SO). 

Ans. =2xAt. mass of Na + At.mass of S+4x At. mass of o 

       = 2x 23.0 + 32 + 4x 16 = 142 

       Mass % of sodium =22×23/142 x 100 32.39% 

Mass o of sulphur= 32/142 x 100=22.53% 

Mass % of oxygen =4×16 / 142 x 100 = 45.07% 

Q. Calculate the amount of carbon dioxide that, would be produced when 

() 1 mole of carbon is burnt in air 

ii) 1mole of carbon is burnt in 16g of dioxygen     

iii) 2 moles of carbon are burnt in 16 g of dioxygen

ANs.  The balanced chemical equation of combustion of carbon in dioxygen(or air) is 

C(s) +     O2  ------------------------------  CO,g) 

1 mol     1 mol                                   1mol

i) In air, carbon will be completely burnt. 1 mol of carbon will give 1 mol of CO, or = 44g 

ii) Since only 16 g of dioxygen is available (0.5 mol), it will combine with only 0.5 mol of carbon. Dioxygen is the limiting reagent. Thus, 0.5 mol of carbon will be burnt to give 0.5 mol of CO, or = 22 g 

iii) In this case also, dioxygen is limiting reagent and only 0.5 mol of carbon will be burnt. It will produce 22 g of CO2

Q. Calculate the mass of sodium acetate required to make 500mL of 0.375 molar aqueous solution. Molar masses of sodium molae aqueous solution. Molar mass of sodium acetate is 82.0245 g mol

ANS. 0.375 M aqueous solution means that 0.375 mol of sodium acetate are present in 1000 mL of solution. 500 mL of the solution should contain sodium acetate   0.375/2 mol

Molar mass of sodium acetate = 82.0245 g mol·

Mass of sodium acetate required =0.375/2 x 82.0245 2 = 15.38 g 

Alternatively, it may be solved as : Mass of sodium acetate / Molar mass Molarity / volume= x 1000 -- 

0.375= Mass of sodium acetate/82.0245/500× 1000

. Mass of sodium acetate = 0.375x82.0245x 500 **** ***** 1000 

=15.38 g

Q. How much copper can be obtained from 100 g copper sulphate (CuSO4) ? (Atomic mass mass of cu= 63.5 amu). 

Ans. 1 mol of CuSO, contains 1 mol (1 gram atom) of Cu 

Molar mass of CuS0, = 63.5 + 32 +4x 16 

                                      = 39.81g

Q. Determine the molecular formula of an oxide of iron in which the mass per cent of iron and ygen are 69.9 and 30.1 respectively. Givenn that the molecular mass of iron oxide is 159.8 and atomic masses: Fe 55.85 amu and 0 =16.00 amu. 

ANS. Calculation of empirical formula =Fe,0, (See Solved example 65

Empirical formula mass of Fe,O =2 x 55.85 +3x 16.00 = 111.7 + 48.00 = 159.7 g mol 

Molecular formula mass =159.8 g mol

                  = 159.8/159.7 = 1 

Molecular formula = (Fe,0,) = Fe,Og 

Q. Calculatethe atomicmass (average) of chlorine 169.7 1 Molecular formula = (Fe,0,) = Fe,Og using the following data: % Natural abundance Molar mas

ANS. Average atomic mass of cl

            = 75.77 × 34.9689 + 24.23 × 36.9659 / 100

           = 35.453

Q. In three moles of ethane (C.H, calculate the following

i) Number of moles of carbon atoms 

ii) Number of moles of hydrogen atoms 

iii) Number of molecules of ethane 

ANS.  i) 1 mole of C,H, contains 2 moles of carbon Ans. Number of moles of carbon in 3 mole of c2H6= 6

ii) 1mole of C,H, contain 6 mole atoms of hydrogen Number of moles of hydrogen atoms in 3 moles of c2H6 = 3×6=18

iii) 1 mole of C,H, = 6.022 x 102 molecules Number of molecules in 3 moles of CHg = 3 x 6.022 x 1023 = 1.807 x 10** molecules 

Q. Pressure is defined as force per unit area. how surface. The SI unit of pressure, pascal is sh.h below 1 Pa = 1Nm2 

If mass of air at sea level is 1034g cm, calculate the pressure in pascal. 

ANS. Pressure is force (i.e., weight) acting per unit But weight = mng ulat Ans. area                       .'. Pressure = Weight per unit area 

= 1034g/cm² × 9.8 ms²× 1kg/1000g ×100cm×100cm/1m×1m × 1N / kgms² × 1 pa/1NM²

= 1.01332×105 pa

Q. What is the SI unit of mass ? How is it defined? 

ANS. The SI unit of mass is kilogram (kg). Kilogram is defined as the mass of platinum-iridium (Pt- block, stored at the International Bureau of Weights and Measures in France. Thus it is the mass of the international prototype of the kilogram.

Q. What do you mean by significant figures? 

ANS. The significant figures in a number are all the certain digits plus one doubtful digit. It depends upon the precision of the scale or instrument used for the measurement. For example, if volume of a liquid is reported to be 18.25 mL, the digits 1, 8 and 2 are certain while 5 is doubtful. So, it has four significant figures (three certain plus one doubtful). 

Q. A sample of drinking water was found to be severely contaminated with chloroform, CHCly supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass 

i) Express this in per cent by mass. 

ii) Determine the molality of chloroform " the water sample.                                                                Ans. i) 15 ppm means that 15 parts of chloroform is present in 106 parts. 

%by mass of CHCI, =15/10  x100 = 1.5 x 10%

ii) molar mass of Chcl = 12+1+3×35.5 = 119.5

Moles of chcl3 = 1.5×10g / 119.5 g mol¹

                              = 1.255× 10

Mass of water = 100g 

Molality = 1.225×10/100 × 1000

                = 1.255×10⁴m

Q. How many significant figures are present in the following 

1. 0.025            2. 208         3. 5005

4. 126,000        5. 500.0       6. 2.0034

ANS. 1. 2            2. 3              3. 4

           4. 3             5. 4          6. 5

Q. Round up the following upto three significant figures

1. 34.216     b 2. 10.4107       3. 0.04597        4. 2808

ANS.  1. 34.2       3. 0.0460

           2. 10.4       6. 2810

Q. Convert the following into basic units

   1. 28.7pm  2. 15.15up    3. 25365mg

ANS. 28.7 PM = 28.7 PM × 10 m/ 1pm = 2.87×10m

        2. 15.15up = 15.15up × 10s/ 1us = 1.515×10 s

        3. 25365 mg = 25365 mg × 1g/ 1000 mg × 1 kg/1000 g

2.5365 ×10 kg

Q. What will be the mass of one C atom in g

ANS. 1 mole of C atoms = 6.022×10 atom =  12g 

6.022 × 10 atoms of C have mass = 12g

1atom of C will have mass = 12/ 6.022×10

                        =1.9927×10g





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