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Some Basic Concepts Of Chemistry

Class 11th Chemistry Part I CBSE Solution
Exercise
  1. Calculate the molar mass of the following: (i) H2O (ii) CO2 (iii) CH4…
  2. Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4).…
  3. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1%…
  4. Calculate the amount of carbon dioxide that could be produced when (i) 1 mole of carbon is…
  5. Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar…
  6. Calculate the concentration of nitric acid in moles per litre in a sample which has a…
  7. How much copper can be obtained from 100 g of copper sulphate (CuSO4)?…
  8. Determine the molecular formula of an oxide of iron in which the mass per cent of iron and…
  9. Calculate the atomic mass (average) of chlorine using the following data:…
  10. In three moles of ethane (C2H6), calculate the following: (i) Number of moles of carbon…
  11. What is the concentration of sugar (C12H22O11) in mol L-1 if its 20 g are dissolved in…
  12. If the density of methanol is 0.793 kg L-1, what is its volume needed for making 2.5 L of…
  13. Pressure is determined as force per unit area of the surface. The SI unit of pressure,…
  14. What is the SI unit of mass? How is it defined?
  15. Match the following prefixes with their multiples:
  16. What do you mean by significant figures?
  17. A sample of drinking water was found to be severely contaminated with chloroform, CHCl3,…
  18. Express the following in the scientific notation: (i) 0.0048 (ii) 234,000 (iii) 8008 (iv)…
  19. How many significant figures are present in the following? (i) 0.0025 (ii) 208 (iii) 5005…
  20. Round up the following up to three significant figures: (i) 34.216 (ii) 10.4107 (iii)…
  21. The following data are obtained when dinitrogen and dioxygen react together to form…
  22. Fill in the blanks in the following conversions: (i) 1 km = ...................... mm =…
  23. If the speed of light is 3.0 × 10^8 m s-1, calculate the distance covered by light in 2.00…
  24. In a reaction Identify the limiting reagent, if any, in the following reaction mixtures.…
  25. Dinitrogen and dihydrogen react with each other to produce ammonia according to the…
  26. How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?
  27. If ten volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes…
  28. Convert the following into basic units: (i) 28.7 pm (ii) 15.15 pm (iii) 25365 mg…
  29. Which one of the following will have largest number of atoms? (i) 1 g Au (s) (ii) 1 g Na…
  30. Calculate the molarity of a solution of ethanol in water in which the mole fraction of…
  31. What will be the mass of one^12 C atom in g?
  32. How many significant figures should be present in the answer of the following…
  33. Use the data given in the following table to calculate the molar mass of naturally…
  34. Calculate the number of atoms in each of the following: (i) 52 moles of Ar (ii) 52 u of He…
  35. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in…
  36. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction,…
  37. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous…

Exercise
Question 1.

Calculate the molar mass of the following:

(i) H2O (ii) CO2 (iii) CH4


Answer: (i) H2O:
The molecular mass of water, H2O
= (2 × Atomic mass of hydrogen) + (1 × Atomic mass of oxygen)
= [2(1.0084) + 1(16.00 u)]
= 2.016 u + 16.00 u
= 18.016
= 18.02 amu

(ii) CO2:
The molecular mass of carbon dioxide, CO2
= (1 × Atomic mass of carbon) + (2 × Atomic mass of oxygen)
= [1(12.011 u) + 2 (16.00 u)]
= 12.011 u + 32.00 u
= 44.01 amu

(iii) CH4:
The molecular mass of methane, CH4
= (1 × Atomic mass of carbon) + (4 × Atomic mass of hydrogen)
= [1(12.011 u) + 4 (1.008 u)]
= 12.011 u + 4.032 u
= 16.043 amu


Question 2.

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


Answer:

Atomic Mass of Sodium = 23g

Atomic Mass of Sulphur = 32g


Atomic Mass of Oxygen = 16g


From the molecular formula of sodium sulphate, it is understood that in 1 mole of sodium sulphate, 1 mole of sulphur, 4 moles of oxygen and 2 moles of sodium is present.


So Molecular Mass of Sodium Sulphate = [23×2] + [32×1] + [16×4]


= 46 + 32 + 64


= 142g


Percentage of Sodium in sodium sulphate = [46×100]/142


= [4600/142]


= 32.394%


Percentage of Sulphur in sodium sulphate = [32×100]/142


= 3200/142


= 22.535%


Percentage of Oxygen in sodium sulphate = [64×100]/142


= 6400/142


=45.070%


Question 3.

Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.


Answer:

Given:

Percentage of Iron in the oxide = 69.9%

Percentage of oxygen in the oxide = 30.1%

To determine the empirical formula of iron oxide, we need to find the simplest whole number ratio.


Simplest molar ratio of iron to oxygen:

= 1.25: 1.88

= 1: 1.5
for converting above ratio in the whole number ratio by 2,we will get

= 2: 3

The mole ratio of the element is represented by subscripts in empirical formula.

Therefore, the empirical formula for iron oxide is Fe2O3.



Question 4.

Calculate the amount of carbon dioxide that could 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 reaction of carbon with oxygen is given as follows:


(i) By observing the above equation, it is understood that when 1 mole of carbon is burnt in air, then the amount of carbon dioxide produced is 44 g [Molecular Mass of carbon dioxide].


(ii) In the above equation 32 g of oxygen react with 1 mole of carbon, then 44 g of carbon dioxide is produced.

32 g of Oxygen → 44 g of Carbon Dioxide
16 g of Oxygen → X g of Carbon Dioxide

X = [44×16]/32

= 22 g

Therefore, 16 g of oxygen when reacted with 1 mole of carbon we get 22 g of CO2.

(iii) Even though 2 moles of carbon is available for reaction but 16 g of oxygen can react with only 0.5 moles of carbon to give 22 g of CO2. In this oxygen acts as limiting reagent and carbon acts as an excess reagent. So 1.5 moles carbon would be left un-reacted.


Question 5.

Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–1.


Answer:

Given:

Molecular Mass of Sodium Acetate = 82.0245g/mol


Molarity of sodium acetate solution required = 0.375 M


0.375 molar solution of sodium acetate means 0.375g of sodium acetate dissolved in 1000ml of the solution.


But it is given in the question that the solution should be of 500ml.


So, 0.375g of CH3COONa → 1000ml of solution


X g of CH3COONa → 500ml of solution


X



= 0.1875 moles


Mass of Sodium acetate required to make 500ml solution = Number of moles of CH3COONa × Molecular Mass of CH3COONa


= 82.0245× 0.1875


= 15.379g


≈ 15.38g


Therefore 15.38g of CH3COONa is required to make 500ml of solution.


Question 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%.


Answer:

Given:

Density of nitric acid = 1.41 g/ml


Percentage of Nitric acid = 69%


69% of nitric acid signifies 69g of nitric acid dissolved in 100g of solution.


Volume of nitric acid solution = [mass of solution]/ density of solution


= 100/1.41


= 70.922 ml


Number of moles of Nitric Acid



= 1.095 moles


Molarity, Mo


= 15.404 M


≈ 15.4 M


Therefore, concentration of nitric acid is 15.4 mole/L.


Question 7.

How much copper can be obtained from 100 g of copper sulphate (CuSO4)?


Answer:

Given:

Mass of Copper Sulphate = 100g


Atomic Mass of Copper = 63.5g


Atomic Mass of Sulphur = 32g


Atomic Mass of Oxygen = 16g


Molecular Mass of CuSO4 = 63.5 + 32 + [4×16]


= 159.5g


63.5g of Copper [Cu] → 159.5g of Copper Sulphate


X g of Copper [Cu] → 100 g of Copper Sulphate


X = [100×63.5]/159.5


= 6350/159.5


= 39.812


≈39.81g


Therefore 39.81g of copper can be obtained from 100g of copper sulphate.



Question 8.

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 molecular mass is 159.69g.


Answer:

Given:

Percentage of Iron in the oxide = 69.9%


Percentage of oxygen in the oxide = 30.1%


Finding Empirical Formula:


To determine the empirical formula of iron oxide, we need to find the simplest whole number ratio.



Therefore, the empirical formula for iron oxide is Fe2O3.


Finding Molecular Formula:


Empirical Formula Mass = [55.84×2] + [16×3]


= 111.68 + 48


= 159.68g


Molecular Mass [given] = 159.69g


n = [Molecular Mass]/ [Empirical Formula Mass]


= 159.69/159.68


= 1


Molecular Formula = n × Empirical Formula


= 1 × Fe2O3.


= Fe2O3


Therefore, the Molecular Formula is Fe2O3.



Question 9.

Calculate the atomic mass (average) of chlorine using the following data:



Answer:

Given:

Atomic Mass of First Isotope of Chlorine = 34.9689g


Natural Abundance of First Isotope of Chlorine, Pa1 = 75.77%


= 0.7577


Total Atomic Mass of 1st Isotope, Ma1 = 34.9689×0.7577


= 26.4959g


Atomic Mass of Second Isotope of Chlorine = 36.9659g


Natural Abundance of Second Isotope of Chlorine, Pa2 = 24.23%


= 0.2423


Total Atomic Mass of 2nd Isotope, Ma2 = 36.9659×0.2423


= 8.9568g

Avg. Atomic Mass of Chlorine =
=(26.4959 gm + 8.9568 gm)

=35.4527 gm.


Question 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.


Answer:

In 1 mole of ethane (C2H6), it is seen that 2 moles of carbon and 6 moles of hydrogen is present.


(i) Since three moles of ethane is present,


So, Number of moles of carbon present = 3×2


= 6


Therefore 6 moles of carbon is present in 3 moles of ethane.


(ii) Since three moles of ethane is present,


So, Number of moles of hydrogen present = 3×6


= 18


Therefore 18 moles of hydrogen is present in 3 moles of ethane.


(iii) 1 mole of ethane contains 6.023×1023 molecules of ethane [By Avogadro’s Law]


Therefore, Number of molecules if ethane in 3 moles of ethane


= 3×6.023×1023


= 18.069×1023


Therefore, Number of molecules of ethane present in 3 moles of ethane is 18.069×1023



Question 11.

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


Answer:

Given:

Mass of sugar = 20g


Volume of solution = 2 L


Atomic Mass of Hydrogen = 1g


Atomic Mass of Oxygen = 16g


Atomic Mass of Carbon = 12g


So Molecular Mass of sugar = [12×12] + [22×1] + [16×11]


= 144 +22 + 176


= 342 g


Number of moles of sugar = [Mass of sugar]/ Molecular Mass of sugar


= 20/342


= 0.05848


Molarity, Mo



= 0.0292 mol/L


≈ 0.029 mol/L



Question 12.

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?


Answer:

Given:

Density of methanol = 0.793 kg/L


Molarity of solution = 0.25 M


Molecular Formula of Methanol = [CH3OH]


Atomic Mass of Hydrogen = 1g


Atomic Mass of Oxygen = 16g


Atomic Mass of Carbon = 12g


So Molecular Mass of methanol= [12×1] + [4×1] + [16×1]


= 12 + 4 + 16


= 32g


= 0.032kg

Molarity =

0.25 =

No. of moles of methanol = 0.625 mole

Weight of methanol = ( 0.625 × 32 )gm

=20 gm

=0.02 kg

Volume of methanol =

=

=0.025 L

Therefore, volume needed for making 2.5 L of its 0.25 M solution is 0.025 L.


Question 13.

Pressure is determined as force per unit area of the surface. The SI unit of pressure, Pascal is as shown below:

1Pa = 1N m–2

If mass of air at sea level is 1034 g cm–2, calculate the pressure in Pascal.


Answer:

Given:


Pressure = Force/Area


=


= 1.01332×105 kg/m/s2


We know that,


1N = 1km/ s2


1 Pa = 1 kg/m-2/s2


= 1 kg/m-1/s2


Therefore, Pressure in Pascal =1.01322×105 Pa



Question 14.

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


Answer:

The SI system was internationally accepted in 1960. SI is the abbreviation of “Systeme International” Units which is French equivalent of “International System of Units”.

The SI unit of Mass is kilogram.


1 kilogram is defined as the mass of a platinum-iridium cylinder kept in Paris.


In practice, the mass of 1litre of water at 4°C is 1 kilogram.


On atomic scale, 1 kilogram is equivalent to the mass of 5.0188× 1025 atoms of 6C12 [isotope of carbon].


Question 15.

Match the following prefixes with their multiples:



Answer:



Question 16.

What do you mean by significant figures?


Answer:

The total number of digits in a measured physical quantity is called the number of significant figures. The number of significant figures refers to the precision of a measured quantity and may be defined as follows:

The number of significant figures in a measured physical quantity refers to number of digits written, including the last one whose value is uncertain.


Rules for Determination of Significant Figures:


[i] All non-zero digits are significant.


For example: 146 cm has three significant figures.


[ii] The zeros placed to the left of the first non-zero digit in the given physical quantity are not significant.


For example: 0.56cm has two significant figures.


[iii] The zeros placed to the right of the decimal point are significant.


For example: 15,0cm has three significant figures.


[iv] The zeros placed between two non-zero digits are significant.


For example: 6.08cm has three significant figures.



Question 17.

A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).

(i) Express this in percent by mass.

(ii) Determine the Molality of chloroform in the water sample.


Answer:

Given:

Level of contamination = 15 ppm [by mass]


To find: Mass Percentage and Molality


Formula:


Molality


Mass Percentage of Solute


Calculation of Mass Percentage:


15 ppm means 15 parts of Chloroform in 1 million [106] parts of drinking water


→Mass Percentage



= 1.5 × 10-3 %


Calculation of Molality:


→Molecular Mass of Chloroform, CHCl3 = [12] + [1] + [35.5×3]


= 119.5 g


→Number of Moles of Chloroform = [15 / 119.5]


= 0.1255 moles


Molality



= 1.255 × 10-4


Therefore, the Mass Percentage is = 1.5 × 10-3% and the Molality of the solution is = 1.255 × 10-4 m.



Question 18.

Express the following in the scientific notation:

(i) 0.0048

(ii) 234,000

(iii) 8008

(iv) 500.0

(v) 6.0012


Answer:

(i) The scientific notation of 0.0048 is 4.8×10-3

(ii) The scientific notation of 234,000 is 2.34×105


(iii) The scientific notation of 8008 is 8.008×103


(iv) The scientific notation of 500 is 5×102


(v) The scientific notation of 6.0012 is 6.0012×100


Question 19.

How many significant figures are present in the following?


(i) 0.0025

(ii) 208
(iii) 5005
(iv) 126,000
(v) 500.0
(vi) 2.0034


Answer:

Rules for Determination of Significant Figures:

[i] All non-zero digits are significant.

For example: 146 cm has three significant figures.

[ii] The zeros placed to the left of the first non-zero digit in the given physical quantity are not significant.

For example: 0.56cm has two significant figures.

[iii] The zeros placed to the right of the decimal point are significant.

For example: 15,0cm has three significant figures.

[iv] The zeros placed between two non-zero digits are significant.

For example: 6.08cm has three significant figures.


(i) 0.0025 – 2 significant figures

(ii) 208 – 3 significant figures


(iii) 5005 – 4 significant figures


(iv) 126,000 – 3 significant figures


(v) 500.0 – 4 significant figures


(vi) 2.0034 – 5 significant figures


Question 20.

Round up the following up to three significant figures:

(i) 34.216

(ii) 10.4107

(iii) 0.04597

(iv) 2808


Answer:

(i) 34.2

(ii) 10.4


(iii)0.0460


(iv)2810



Question 21.

The following data are obtained when dinitrogen and dioxygen react together to form different compounds:



(a) Which law of chemical combination is obeyed by the above experimental data?

Give its statement.


Answer:

The Law satisfied by the above data is named as Law of Multiple Proportions. This law was given by John Dalton in 1804.

Statement:


When two elements combine to form two or more compounds, then the different masses of one element, which combine with a fixed mass of the other element, bear a simple whole number ratio with another.


Nitrogen and oxygen combine together to form five compounds- Nitrous Oxide [N2O], Nitric Oxide [NO], Nitrogen Trioxide [N2O3], Nitrogen Dioxide [N2O4] and Nitrogen pentoxide [N2O5]. In these compounds ratio of nitrogen and oxygen by mass is given in the above table.


Thus the ratio of the masses of oxygen which combine with fixed mass of nitrogen are 16:32:48:64:80 that is they are in simple whole number ratio of 1:2:3:4:5.



Question 22.

Fill in the blanks in the following conversions:

(i) 1 km = ...................... mm = ...................... pm

(ii) 1 mg = ...................... kg = ...................... ng

(iii) 1 mL =...................... L = ...................... dm3


Answer:

(i) We know the basic relation that

1km = 1000m


1 m = 1000 mm


Therefore 1 km = 1000×1000


= 106 mm


We also know that


1 m = 1012pm


Therefore 1 km = 1000× 1012 pm


= 1015pm


(ii) We know the basic relation that


1 kg = 1000 g


1 g = 1000 mg


So, 1 mg = 10-3g


Therefore 1 mg = 10-3 × 10-3 kg


= 10-6 kg


1 g = 109ng


1000 mg = 109 ng


1 mg = [109]/103 ng


= 106 ng


(iii) We know the basic relation that


1 L = 1000 ml


∴ 1 ml = 10-3 L


1 ml = 10-3 dm3



Question 23.

If the speed of light is 3.0 × 108 m s–1, calculate the distance covered by light in 2.00 ns.


Answer:

Given:

Speed of light = 3.0 × 108 m s–1


Time in which distance is to be covered = 2.00 ns


Speed = [Distance]/ Time


Distance = [3.0 × 108] × [2×10-9]


= 0.6 m



Question 24.

In a reaction



Identify the limiting reagent, if any, in the following reaction mixtures.

(i) 300 atoms of A + 200 molecules of B

(ii) 2 mol A + 3 mol B

(iii) 100 atoms of A + 100 molecules of B

(iv) 5 mol A + 2.5 mol B

(v) 2.5 mol A + 5 mol B


Answer:

While solving the problems relating to chemical equation, the reactant reacts according to the balanced chemical equation. Quite often, one of the reactant is present in lesser amount while the other may be present in higher amount. The reactant which is present in lesser amount is known as limiting reactant or reagent.


[i] According to the above given reaction, 1 atom of A reacts with 1 molecule of B. Thus, 200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unused. Hence, B is the limiting reagent as atom B is in lesser amount [200].


[ii] According to the reaction given in the question, 1 mol of A reacts with 1 mol of B. Thus, 2 mol of A will react with only 2 mol of B. As a result, 1 mol of A is not being consumed in the reaction process. Hence, A is the limiting reagent.


[iii] According to the given reaction given in the question, 1 atom of A combines with 1 molecule of B. Thus, all 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric where no limiting reagent is present.


[iv] 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of B will combine with only 2.5 mol of A. As a result, 2.5 mol of A will be left as such as it is not consumed in the reaction. Hence, B is the limiting reagent because B is less as compared to A.


[v] According to the reaction in the question, 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be left as such. Hence, A is the limiting reagent as it is present in lesser amount.



Question 25.

Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:



(i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 ×103 g of dihydrogen.

(ii) Will any of the two reactants remain unreacted?

(iii) If yes, which one and what would be its mass?


Answer:

The reaction given in the question is not balanced.

Balancing the above chemical reaction:


N2 + 3H2→ 2NH3


(i) Finding mass of NH3 produced:


Atomic Mass of Hydrogen = 1g


Atomic Mass of Nitrogen = 14g


Molecular Mass of NH3 = 14 + 3


= 17 g

N2 + 3H2→ 2NH3
28gm 6gm 34gm

From the chemical equation we observe that 28g of nitrogen react with 6g of hydrogen to give 34g of ammonia


Therefore 1 mole of dinitrogen reacts with 3 moles dihydrogen to give 2 moles of ammonia.


Therefore, amount of dihydrogen required to react with 2000g of dinitrogen = [6×2000]/28


= 12000/28


= 428.57g


Therefore, 2000g of dinitrogen will produce = [34×2000]/28


= 68000/28


= 2428.57g of Ammonia


Therefore 2428.57g of ammonia is produce when 2000g of dinitrogen is used.


(ii) Since dinitrogen is present in lesser amount(428.57 gm), it is the limiting reagent and dihydrogen is the excess reagent and hence H2 will be left unreacted.


(iii) Mass of dihydrogen left unreacted = 1000 – 428.57


= 571.43g


Question 26.

How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?


Answer:

Atomic Mass of Oxygen = 16g

Atomic Mass of Carbon = 12g


Atomic Mass of Sodium = 23g


Molecular Mass of Sodium Carbonate = [23×2] + [12×1]+[16×3]


= 46 + 12 + 48


= 106g


1 mole of Na2CO3 = 106 g


0.5 mole of Na2CO3 = X g


X = [0.5×106]/1


= 53g of Na2CO3


0.5M of Na2CO3 signifies 0.5 moles of Na2CO3 dissolved in 1 litre of water or rather 53g of Na2CO3 dissolved in 1 litre of water.



Question 27.

If ten volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapour would be produced?


Answer:

The reaction of hydrogen with oxygen is given as below:

2H2 + O2→ 2H2O


By referring to above equation, 2 volumes of hydrogen and 1 volume of oxygen react together to form 2 volumes of water.


So if 10 volumes of hydrogen react with 5 volumes of oxygen then 10 volumes of water is produced.



Question 28.

Convert the following into basic units:

(i) 28.7 pm

(ii) 15.15 pm

(iii) 25365 mg


Answer:

(i) 1 pm = 10-12 m

28.7 pm = 28.7×10-12m


= 2.87×10-11m


(ii) 1 pm = 10-12 m


15.15 pm = 15.15×10-12m


= 1.515×10-11m


(iii) 1 mg = 10-3 g


1 mg = 10-6 kg


∴ 25365 mg = 25365×10-6 kg


= 2.5365×10-2 kg


Question 29.

Which one of the following will have the largest number of atoms?
(i) 1 g Au (s)

(ii) 1 g Na (s)

(iii) 1 g Li (s)

(iv) 1 g of Cl2 (g)


Answer:

(i) Gram atomic mass of Au= 197 g

Or

197g of Au contains = 6.022 x 1023

Therefore 1gm of Au contains = 6.022 x 1023/197X1 = 3.06 x 1021atoms

(ii) Gram atomic mass of Na = 23 g

Or

23 g of Na contains atoms = 6.022x 1023

Or

1gm of Na contains atoms = 6.022x1023/23 X1 = 26.2 x1021atoms

(iii) Gram atomic mass of Li = 7

Or

7g of Li contains atoms = 6.022 x 1023

Or

1g of Li contains atoms = 6.022 x 1023/7 X1= 86.0 x 1021atoms

(iv) Gram atomic mass of Cl = 71 Or 71g of Cl contains atoms = 6.022x1023

Or

1 g of Cl contains atoms = 6.022x1023/71 X 1= 8.48 x 1021atoms

Therefore, 1 g of Li will have the largest number of atoms.


Question 30.

Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).


Answer:

Given:

Mole Fraction of Ethanol = 0.040


Density of water = 1g/ml(given)


So, Mole fraction of water = 1- 0.04 = 0.96

Let,total mole in the solution=1

So, Moles of Ethanol = 0.04 moles.

Therefore, mass of 0.04 moles of ethanol = 0.04 mole × 46 gm/mole = 1.86 gm.

Density of ethanol = 0.789 g/ml.

Volume of ethanol = = 2.357 ml = 0.0023 L

Now, moles of water= 1-0.04 = 0.96 moles.

Mass of water in this solution = 0.96 mole × 18 gm/mol = 17.28 gm.

Volume of water = =17.28 ml = 0.017 L.

Molarity of Solution = ==2.043 mol/L.



Question 31.

What will be the mass of one 12C atom in g?


Answer:

By Avogadro’s Law

1 mole of carbon contain 6.023×1023 carbon atoms = 12g of Carbon


6.023×1023 atoms of carbon → 12g of Carbon


1 atom of Carbon → X g of Carbon


X = [12]/ 6.023×1023


= 1.992×10-23g of Carbon


Therefore, mass of 1 atom of Carbon is 1.992×10-23g.



Question 32.

How many significant figures should be present in the answer of the following calculations?

(i)

(ii)

(iii) 0.0125 + 0.7864 + 0.0215


Answer:

(i) = 1.648


Least precise number of calculation = 298.15


Number of significant numbers in result = Number of significant figures in least precise number = 2


Correct Answer = 1.65


(ii) Calculated Answer = 26.82


Least precise number of calculation = 5.364


Number of significant numbers in result = Number of significant figures in least precise number = 3


Correct Answer = 26.820


(iii) Calculated Answer = 0.8204


Number of significant numbers in result = Number of significant figures in least precise number = 4


Correct Answer = 0.8204



Question 33.

Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:



Answer:

Atomic Mass of First Isotope of Argon = 35.96755g

Natural Abundance of First Isotope of Argon, Pa1= 0.337% = 0.00337


Total Atomic Mass of 1st Isotope, Ma1 = 35.96755×0.00337


= 0.12121g


Atomic Mass of Second Isotope of Argon = 37.96272g


Natural Abundance of Second Isotope of Argon, Pa2 = 0.063%


=0.00063


Total Atomic Mass of 2nd Isotope, Ma2 = 37.96272×0.00063


= 0.02392g


Atomic Mass of Third Isotope of Argon = 39.9624g


Natural Abundance of Third Isotope of Argon, Pa3 = 99.6%


Total Atomic Mass of 3rd Isotope, Ma3 = 39.9624×0.996


= 39.8026g


So Average Atomic Mass of Argon =



→ 39.9477g


Therefore, the average atomic mass of Argon is 39.95g



Question 34.

Calculate the number of atoms in each of the following:

(i) 52 moles of Ar

(ii) 52 u of He

(iii) 52 g of He.


Answer:

(i) According to Avogadro’s Law,

1 mole of Argon → 6.023×1023 atoms of Argon


So, 52 moles of Argon → X atoms of Argon


X =52×6.023×1023


=3.13196×1025


Therefore in 52 moles of Argon 3.13196×1025 atoms are present.


(ii) Atomic Mass of Helium [He] = 4 u

This means that mass of one atom of Helium is = 4 u

So, Number of atom of Helium contained in 52 u = 52/4= 13 atoms

Therefore, in 52 u of Helium 13 atoms are present.


(iii) Atomic Mass of Helium [He] = 4 u

Number of Moles of Helium in 52 g = 52/4= 13 moles

So, Number of atoms in 52g of Helium [He] = 13×6.023×1023= 7.8299×1024


Therefore, in 52 g of Helium 7.83×1024 atoms are present.


Question 35.

A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate

(i) empirical formula,

(ii) molar mass of the gas

(iii) molecular formula.


Answer:

Given:

Mass of Carbon = 3.38g


Volume of welding gas = 10 L


Mass of 10 L of welding gas = 11.6g


Finding Percentage of Carbon:


44 parts of CO2→12 parts of C


OR


44g of CO2→12 g of C


Therefore, according to the question


3.38 g of CO2→ X g of C


X = [12× 3.38] /44


=0.921g


18 g of water → 2g of hydrogen


So, 0.690 g of water → X g of hydrogen


X = [2× 0.690]/18


=0.0767g


Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is obtained as follows:


=0.9217 g + 0.0767 g = 0.9984 g


Percentage of carbon = [100×weight of carbon]/weight of compound


= [0.921× 100]/0.998


= 92.32%


Also percentage of hydrogen =[100×weight of hydrogen]/weight of compound


=[0.0766× 100]/0.998


=7.68%


(i) Finding Empirical Formula:



Empirical formula of the compound = CH


Therefore, the Empirical Formula of the compound is CH


(ii) Weight of 22.4 L of gas of STP = [11.6g×22.4L]/10.0L


= 25.985g


≈ 26 g


Therefore, the molecular mass of the gas is 26 g.


(iii) Finding Molecular formula:


Empirical formula mass = 12+1 = 13g


Also molecular mass =26 g [as obtained in step (ii)]


n = [Molecular Mass] / [Empirical Formula Mass]


=26/13


=2


Now molecular formula = n ×Empirical Formula = 2×CH = C2H2


Therefore, the molecular formula of the compound is C2H2.



Question 36.

Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction,



What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?


Answer:

Given:

Volume of HCl = 25 ml


Molarity of HCl = 0.75 M


Atomic Mass of Hydrogen = 1g


Atomic Mass of Chlorine = 35.5g


Molecular Mass of HCl = 1 + 35.5


= 36.5g


By Avogadro’s Law,


1 mole of HCl→ 36.5g of HCl


So 0.75 moles of HCl → X g of HCl


X = 0.75×36.5


X = 27.35 g


Thus, 1000 mL of solution contains HCl = 27.375g


1000ml of HCl → 27.357g of HCl


So 25 ml of HCl → X g of HCl


X = [25×27.375]/1000


= 684.375/1000


=0.68438 g


From the chemical equation given in the question,


CaCO3+2 HCL →CaCL2+H2O


We observe that


2 moles of HCl reacts with 1 mole of CaCO3


73 g of HCl reacts with 100g of CaCO3


0.68438g of HCl reacts with X g of CaCO3


X = [0.68483×100]/73


= 68.483/73


= 0.93812 g


So amount of CaCO3 required is 0.94g.



Question 37.

Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction



How many grams of HCl react with 5.0 g of manganese dioxide?


Answer:

Given:

Mass of Manganese Dioxide = 5 g


Atomic Mass of Manganese = 55g


Atomic Mass of Oxygen = 16g


Atomic Mass of Hydrogen = 1g


Molecular Mass of MnO2 = 55 + [16×2]


= 55 + 32


= 87g


Molecular Mass of HCl = 1 + 35.5


= 36.5g


So by observing the chemical reaction we see that,


1moles of MnO2 reacts with 4moles of HCl


87g of MnO2→ 4×36.5g of HCl


So 5g of MnO2→ X g of HCl


X = [146×5]/87


X = 730/87


X = 8.39g


So, 8.4g of HCl will react completely with 5g of MnO2.