Ncert Solution of Chapter Basic Concepts Of chemistry

Q. 1. Calculate the molecular mass of the following : ( i ) H₂O ( ii ) CO₂ ( iii ) CH4

( i ) Molecular mass of H₂O

= 2 ( 1.008u ) + ( 16.00u ) = 18.016 u

Molecular mass of CO₂

= 1 ( 12.01 u ) + 2 ( 16.00u ) = 44.01u

( iii ) Molecular mass of CH4

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

Q2.Calculate the mass per cent of different elements present in sodium sulphate ( Na₂SO4 ) .

= Molar mass of Na₂SO4 = 2 x At . mass of Na + atomic mass of S + 4 × atomic mass of O

= 2 x 23.0 + 32 + 4 x 16 = 142

Mass % of sodium =2 × 23/142 x 100

= 32.39 %

Mass % of sulphur = 32/142 x 10

=22.53%

Mass of oxygen = 4x16 /142 x 100

= 45.07 %

Q3. Determine the empirical formula of an oxide of iron which has 69.9 % iron and 30.1 % dioxygen by mass . ( Atomic masses : Fe = 55.85 amu , 0 = 16.00 amu )

Q4.Calculate the amount of carbondioxide that would be produced when

( i ) 1 mole of carbon is burnt in air

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

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

Answer :- The balanced chemical equation of combustion of carbon in dioxygen ( or air) is -

C ( s ) + O₂(g) ―> CO₂

1 mol 1 mol 1 mol

( i ) In air , carbon will be completely burnt. 1 mol of carbon will give 1 mol of CO₂ 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 CO₂ 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 .

Q5.Calculate the mass of sodium acetate

( CH₂COONa ) required to make 500 mL of 0.375 molar aqueous solution . Molar mass of sodium acetate is 82.0245g mol¹.

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 =

82.0245g mol - ¹

Mass of sodium acetate required =

0.375 x 82.0245 / 2 = 15.38g

Q. 6. Calculate the concentration of nitric acid in moles per litre in a sample which has a density , 1.41 g mL - ¹ and the mass per cent of nitric acid in it being 69 % .

Ans . 69 mass percent of nitric acid means that 69 g of HNO3 are present in 100 g if solution.

Volume of solution= Mass/Density

= 100/1.41 g mL–1

= 70.92 mL

Moles of HNO3 = 69/63

Therefore ,

Molarity = Moles of HNO3 / Volume of solution (in mL) ×1000

= 15.44 M

Q. 7. How much copper can be obtained from 100 g of copper sulphate ( CuSO ) ? ( Atomic mass of Cu = 63.5 amu ) .

Ans :- 1 mol of CuSO4 contains 1 mol ( 1 gram atom ) of Cu Molar mass of CuSO4 = 63.5+ 32 + 4 × 16 = 159.5 g

Therefore, 159.5 g of CuSO4 will give Cu

= 63.5 g

100 g of CuSO4 will give Cu = 63.5/15.95×100 = 39.81 g

Q8.Determine the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1 respectively . Given 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 = Fe₂O3 .

Empirical formula mass of Fe₂O3 = 2 × 55.85 + 3 × 16.00

= 111.7 + 48.00

= 159.7 g mol-1

Molecular formula mass = 159.8 g mol-1

Divide molecular formula mass by empirical formula mass

n = molecular formula mass / Empirical formula mass

= 159.8 / 159.7 = 1

Therefore, Molecular Formula = (Fe₂O₂) = Fe₂O3

Q. 9. Calculate the atomic mass ( average) of chlorine using the following data
% Natural abundance Molar mass

35Cl 75.77. 34.9689

37CI 24.23. 36.9659

Ans. Average atomic mass of Cl =
75.77x34.9689 +24.23x36.9659 /100

= 35.453

Q. 10. In three moles of ethane ( C2H6), 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 C2H6 contains 2 moles of carbon
:. Number of moles of carbon in 3 moles of
C₂H6= 6
(ii) 1 mole of C₂H6 contain 6 mole atoms of hydrogen :

:. number of moles of hydrogen atoms in 3 moles of C2H6 = 3×6 = 18

(iii) 1 mole of C2H6 = 6.022 x 10³ molecules
Number of molecules in 3 moles of
C₂H6 = 3 x 6.022 x 10 to the power 23= 1.807 x 1024 molecules

Q. 11. What is the concentration of sugar (C12H22O11)in mol L-¹ if its 20g are dissolved in enough water to make a final volume upto 2 L ?

Ans.

Q12. If the density of methanol is 0.793 kg L-¹, what is its volume needed for making 2.5 L of its 0.25 M solution ?

Ans :-

Q13. Pressure is defined as force per unit area of the surface. The SI unit of pressure, pascal is shown below:

1 Pa = 1Nm²
If mass of air at sea level is 1034 g cm2, calculate the pressure in pascal.
Ans :- Pressure is force (i.e., weight) acting per unit area
But weight = mg
:. Pressure = Weight per unit area
=1034/cm2 × 9.8 ms-2

=1034g/cm2 × 9.8 ms-2 × 1 kg /1000 g × 100 cm x 100cm / 1m × 1m × 1N/ kg ms-2 × 1Pa/1Nm-2= 1.01332 x 10³ Pa

Q. 14. 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-Ir) 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. 15. Match the following prefixes with their multiples
Ans.

Q16. What do you mean by significant figures ?
Ans.The significant figures in a number are all the certain digits plus ane 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. 17. A sample of drinking water was found to be severely contaminated with chloroform, CHCI3,supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).

(1) Express this in per cent by mass.
(ii) Determine the molality of chloroform in
the water sample.
Ans. (i) 15 ppm means that 15 parts of chloroform is present in 10 power 6 parts.
:. %by mass of CHCI3 = 15 / 10 to the power 6 x100 = 15 x 10-3%
ii) Molar mass of CHCl3 = 12+1+3x 35.5 = 119.5

Moles of CHCl3 =1.5×10-3 / 119.5g mol-1

= 1.255 x 10-5

Mass of water = 100 g

:. Molality = 1.225 × 10 -4 m

Q. 18. Express the following in the scientific notation.
(i) 0.0048 (ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012

Ans.