Chapter 1 Some Basic Concepts of Chemistry

Answer the following Questions.

1. Calculate the molecular mass of the following :

(i) H₂O

(ii) CO₂

(iii) CH₄

Solution :

(i)CH₄ :

Molecular weight of methane, CH₄

= (1 x Atomic weight of carbon) + (4 x Atomic weight of hydrogen)

= [1(12.011 u) +4 (1.008u)]

= 12.011u + 4.032 u

= 16.043 u

(ii) H₂O :

Molecular weight of water, H₂O

= (2 x Atomic weight of hydrogen) + (1 x Atomic weight of oxygen)

= [2(1.0084) + 1(16.00 u)]

= 2.016 u +16.00 u

= 18.016u

So approximately

= 18.02 u

(iii) CO₂ :

= Molecular weight of carbon dioxide, CO₂

= (1 x Atomic weight of carbon) + (2 x Atomic weight of oxygen)

= [1(12.011 u) + 2(16.00 u)]

= 12.011 u +32.00 u

= 44.011 u

So approximately

= 44.01u

2. Calculate the mass percent of different elements present in Sodium Sulphate (Na₂SO₄).
Solution :

Molar mass of Na₂SO₄ = [(2 × 23.0) + (32.00) + 4 (16.00)] = 142 g

Mass percent of an element = (Mass of that element in compound/Molar mass of that compound) × 100

∴ Mass percent of sodium (Na): (46/142) × 100 = 32.39%

Mass percent of sulphur(S): (32/142) × 100 = 22.54%
Mass percent of oxygen:(O): (64/142) × 100 = 45.07%

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

Solution :

% of iron by mass = 69.9 % [Given]
% of oxygen by mass = 30.1 % [Given]

Atomic mass of iron = 55.85 amu

Atomic mass of oxygen = 16.00 amu
Relative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25
Relative moles of oxygen in iron oxide = %mass of oxygen by mass/Atomic mass of oxygen = 30.01/16=1.88

Simplest molar ratio = 1.25/1.25 : 1.88/1.25 

 ⇒ 1 : 1.5 = 2 : 3

∴ The empirical formula of the iron oxide is Fe₂O₃.

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.
Solution :
The balanced reaction of combustion of carbon in dioxygens is:
C(s)     +    O2(g)     →    CO₂  (g)
1mole     1mole(32g)     1mole(44g)

(i) In dioxygen, combustion is complete. Therefore 1 mole of carbon dioxide  produced by burning 1 mole of carbon.
(ii) Here, oxgen acts as a limiting reagent as only 16 g of dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide.
(iii) Here again oxgen acts as a limiting reagent as only 16 g of dioxygen is available. It is a limiting reactant. Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.

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 82.0245g mol^-1.
Solution :
0.375 M aqueous solution of sodium acetate means that 1000 mL of solution containing 0.375 moles of sodium acetate.

∴No. of moles of sodium acetate in 500 mL = (0.375/1000)×500 = 0.375/2 = 0.1875

Molar mass of sodium acetate = 82.0245g mol^-1

∴Mass of sodium acetate acquired =  0.1875×82.0245 g = 15.380g

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%.
Solution :
Mass percent of 69% means tat 100g of nitric acid solution contain 69 g of nitric acid by mass.
Molar mass of nitric acid (HNO3) = 1+14+48 = 63g mol^-1
Number of moles in 69 g of HNO3 = 69/63 moles = 1.095 moles
Volume of 100g  nitric acid solution = 100/1.41 mL = 70.92 mL = 0.07092 L
∴ Conc. of HNO₃ in moles per litre = 1.095/0.07092 = 15.44 M

7. How much copper can be obtained from 100 g of copper sulphate (CuSO₄ )?
Solution :
1 mole of CuSO₄ contains 1 mole of copper.
Molar mass of CuSO₄ = (63.5) + (32.00) + 4(16.00)
= 63.5 + 32.00 + 64.00 = 159.5 g
159.5 g of CuSO₄ contains 63.5 g of copper.
∴ copper can be obtained from 100 g of copper sulphate = (63.5/159.5)×100 = 39.81g

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 the molar mass of the oxide is 159.69 g mol^-1
Solution :
% of iron by mass = 69.9 % [Given]
% of oxygen by mass = 30.1 % [Given]

Atomic mass of iron = 55.85 amu

Atomic mass of oxygen = 16.00 amu
Relative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85  = 1.25
Relative moles of oxygen in iron oxide = %mass of oxygen by mass /Atomic mass of oxygen = 30.01/16 =1.88

Simplest molar ratio = 1.25/1.25  : 1.88/1.25 

⇒  1 : 1.5 = 2 : 3

∴ The empirical formula of the iron oxide is Fe₂O₃.
Mass of Fe₂O₃ = (2×55.85) + (3×16.00) = 159.7 g mol^-1
n = Molar mass /Empirical formula mass = 159.7/ 159.6 = 1(approx)

Thus, Molecular formula is same as Empirical Formula i.e.  Fe₂O₃.

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

                  % Natural Abundance             Molar Mass

³⁵Cl                      75.77                              34.9689

³⁷Cl                      24.23                              36.9659
Solution :
Fractional Abundance of ³⁵Cl =  0.7577 and Molar mass = 34.9689
Fractional Abundance of ³⁷Cl =  0.2423 and Molar mass = 36.9659
∴ Average Atomic mass = (0.7577 × 34.9689) amu + (0.2423 × 36.9659)
= 26.4959 + 8.9568 = 35.4527

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.
Solution :
(i) 1 mole of C₂H₆  contains 2 moles of Carbon atoms
∴ 3 moles of of C₂H₆  will contain 6 moles of Carbon atoms
(ii) 1 mole of C₂H₆  contains 6 moles of Hydrogen atoms
∴ 3 moles of of C₂H₆  will contain 18 moles of Hydrogen atoms
(iii) 1 mole of C₂H₆  contains Avogadro’s no. 6.02 ×10²³ molecules
∴ 3 moles of of C₂H₆  will contain ethane molecule = 3×6.02 × 10²³= 18.06 ×10²³  molecules.

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?
Solution :
Molar mass of sugar (C₁₂H₂₂O₁₁)  = (12 ×12) +(1 ×22)+ (11×16) = 342 g mol^-1
No. of moles in 20g of sugar = 20/342 = 0.0585 mole

Volume of Solution = 2L (given)

Molar concentration = Moles of solute/Volume of solution in L = 0.0585mol /2L = 0.0293 mol L-1 = 0.0293 M

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?
Solution :
Molar mass of methanol (CH₃OH) = (1×12) + (4×1) + (1×16) = 32 g mol-1 = 0.032 kg mol^-1
Molarity of the solution = 0.793 /0.032 = 24.78 mol L^-1 

Applying, M₁V₁ (Given Solution) = M₂V₂ (Solution to be prepared)
24.78×V₁ = 0.25×2.5 L
V₁= 0.02522 L = 25.22 mL

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.
Solution :
Pressure is the force (i.e. weigh) acting per unit area.
P= F/A = 1034g × 9.8ms ^-2/cm²
  =  1034g  × 9.8ms^-2 /cm² × 1kg /1000g × 100cm  /1m × 100cm /1m = 1.01332 ×105 N
Now,
1Pa = 1N m^-2
∴ 1.01332 × 10⁵  N ×m^-2 = 1.01332 ×10⁵  Pa

14. What is the SI unit of mass? How is it defined?
Solution :
The SI unit of mass is kilogram (kg).

The kg is defined as the mass of platinum-iridium (Pt-Ir) cylinder that is stored in an air-tight jar at International Bureau of Weigh and Measures in France.

15. Match the following prefixes with their multiples:

	Prefixes	Multiples
(a)	femto	10
(b)	giga	10−15
(c)	mega	10−6
(d)	deca	109
(e)	micro	106

Solution :

	Prefixes	Multiples
(a)	femto	10−15
(b)	giga	109
(c)	mega	106
(d)	deca	10
(e)	micro	10−6

16. What do you mean by significant figures ?

Solution :

Significant figures are meaningful digits which are known with certainty including the last digit whose value is uncertain.

For example,

In 11.2546 g, there are 6 significant figures but here 11.254 is certain and 6 is uncertain and the uncertainty would be ±1 in the last digit. Hence last uncertain digit is also included in Significant figures.
 

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.
Solution :
(i) 15 ppm means 5 parts in million(10⁶) parts.
∴ % by mass = 15/10⁶ × 100 = 15 × 10^-4 = 1.5×10^-3 %
(ii) Molar mass of chloroform(CHCl³) = 12+1+ (3×35.5) = 118.5 g mol^-1
100g of the sample contain chloroform = 1.5×10-3g
∴ 1000 g(1 kg) of the sample will contain chloroform = 1.5×10^-2 g
= 1.5×10^-2/ 118.65 mole = 1.266 ×10^-4 mole
∴ Molality = 1.266×10^-4 m.

18. Express the following in the scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012
Solution :
(i) 0.0048 = 4.8× 10^-3
(ii) 234, 000 = 2.34× 10⁵
(iii) 8008 = 8.008× 10³
(iv) 500.0 = 5.000× 10²
(v) 6.0012 = 6.0012× 10⁰

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

Solution :

(i) 2
(ii) 3
(iii) 4
(iv) 3
(v) 4
(vi) 5

20. Round up the following upto three significant figures:
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808

Solution :

(i) 34.2
(ii) 10.4
(iii) 0.046
(iv) 2810

21. The following data are obtained when dinitrogen and dioxygen react together to form different compounds:
 Mass of dinitrogen          Mass of dioxygen
(i)14 g                                      16 g
(ii)14 g                                      32 g
(iii)28 g                                      32 g
(iv)28 g                                      80 g
(a) Which law of chemical combination is obeyed by the above experimental data?Give its statement.
(b) Fill in the blanks in the following conversions:
(i) 1 km = …………………. mm = …………………. pm
(ii) 1 mg = …………………. kg = …………………. ng
(iii) 1 mL = …………………. L = …………………. dm³
Solution :
(a) Fixing the mass of dinitrogen as 28 g, masses of dioxygen combined will be 32, 64, 32 and 80 g in the given four oxides. These masses of dioxygen bears a simple whole number ratio as 2:4:2:5. Hence, the data given will obey the law of multiple proportions.
The statement is as follows two elements always combine in  a fixed mass of other bearing a simple ratio to another to form two or more chemical compounds.
(b) (i) 1 km =  1km× 1000m/ 1km ×100cm /1m/ 10mm /1cm = 10⁶ mm
1 km =  1km× 1000m / 1km × 1pm/ 10^-12m = 10¹⁵ pm
(ii) 1 mg = 1mg ×1g/ 1000mg × 1kg / 1000g = 10-6 kg
1 mg = 1mg ×1g/ 1000mg × 1ng/ 10^-9g = 10^-6 ng
(iii) 1 mL = 1mL×1L/ 1000mL = 10^-3 L
1 mL = 1cm³ = 1cm³× (1dm × 1dm × 1dm/ 10cm × 10cm × 10cm) = 10³dm³

22. If the speed of light is 3.0 × 108ms-1, calculate the distance covered by light in 2.00 ns.
Solution :

Distance covered = Speed  × Time = 3.0 × 108ms^-1 × 2.00 ns
= 3.0 × 108ms^-1 × 2.00 ns ×10^-9s  /1ns = 6.00×10^-1m = 0.600m

23. In a reaction
A + B₂ → AB₂
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
Solution :
(i) According to the reaction, 1 atom of A reacts with 1 molecule of B.

 ∴200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unreacted. Hence, B is the limiting reagent.

(ii) According to the reaction, 1 mol of A reacts with 1 mol of B. 

∴ 2 mol of A will react with only 2 mol of B leaving 1 mol of B. Hence, A is the limiting reagent.

(iii) 1 atom of A combines with 1 molecule of B.

∴ All 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric and ther is no limiting reagent.

(iv) 1 mol of atom A combines with 1 mol of molecule B. 

∴ 2.5 mol of B will combine with only 2.5 mol of A. and 2.5 mol of A will be left unreacted. Hence, B is the limiting reagent.

(v) 1 mol of atom A combines with 1 mol of molecule B. 

∴ 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. Hence, A is the limiting reagent.

24. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
N₂(g) + H₂(g) → 2NH₃(g)
(i) Calculate the mass of ammonia produced if 2.00×10³g dinitrogen reacts with 1.00×10³g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
Solution :

1 mole of dinitrogen (28g) reacts with 3 mole of dihydrogen (6g) to give 2 mole of ammonia (34g).

∴ 2000 g of N₂ will react with H₂ = 6/28 × 200g = 428.6g. Thus, here N₂ is the limiting reagent while H₂ is in excess.

28g of N₂ produce 34g of NH₃.

∴2000g of N₂ will produce = 34/28 × 2000g = 2428.57 g of NH₃.

(ii) N₂ is the limiting reagent and H2 is the excess reagent. Hence, H₂ will remain unreacted.

(iii) Mass of dihydrogen left unreacted = 1000g – 428.6g = 571.4 g

25. How are 0.50 mol Na₂CO₃ and 0.50 (M) Na₂ CO₃  different?
Solution :
Molar mass of Na₂CO₃ = (2×23) +12.00+(3×16) = 106 g mol^-1

∴0.50 mol Na₂CO₃ means 0.50 ×106g = 53g

0.50 M Na2CO3 means 0.50 mol of Na2CO₃ i.e. 53g of  Na₂CO₃ are present in 1litre of the solution.

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

Solution :

Dihydrogen gas reacts with dioxygen gas as, 

2H₂(g) + O₂(g) → 2H₂O(g)

Thus, two volumes of dihydrogen react with one volume of dihydrogen to produce two volumes of water vapour. Hence, ten volumes of dihydrogen will react with five volumes of dioxygen to produce ten volumes of water vapour.

27. Convert the following into basic units:
(i) 28.7 pm
(ii) 15.15 pm
(iii) 25365 mg

Solution :

(i) 1 pm = 10^-12 m
28.7 pm = 28.7 × 10^-12 m = 2.87 × 10^-11 m
(ii) 1 pm = 10^-12 m
∴15.15 pm = 15.15 × 10^-12 m = 1.515 × 10^-11 m
(iii) 1 mg = 10^-3 g
25365 mg = 2.5365 × 104×10^-3 g
Now,
1 g = 10^-3 kg
2.5365×10 g = 2.5365 × 10×10^-3 kg
∴25365 mg = 2.5365 × 10^-2 kg

28. Which one of the following will have 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)
Solution :
(i) 1 g Au = 1/197 mol = 1/197 × 6.022×10²³ atoms
(ii) 1 g Na = 1/23 mol = 1/23 × 6.022×10²³ atoms
(iii) 1 g Li = 1/7 mol = 1/7 × 6.022×10²³ atoms
(iv) 1 g Cl2 = 1/71 mol = 1/71 × 6.022×10²³ atoms
Thus, 1 g of Li has the largest number of atoms.

29. 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).

Solution :
Mole fraction of C₂H₅OH = No. of moles of C₂H₅OH /No. of moles of solution
nC₂H₅OH = n(C₂H₅OH) / (C₂H₅OH) + n(H₂O) = 0.040 (Given) … 1
We have to find the number of moles of ethanol in 1L of the solution but the solution is dilute. Therefor, water is approx. 1L.
No. of moles in 1L of water = 1000g  /18g mol^-1 = 55.55 moles
Substituting n(H₂O) = 55.55 in equation 1
n(C2H5OH) / (C₂H₅OH) + 55.55 = 0.040
⇒ 0.96n (C₂H₅OH) = 55.55 × 0.040
⇒ n(C₂H₅OH) = 2.31 mol
Hence, molarity of the solution = 2.31M

30. What will be the mass of one  12C atom in g ?
Solution :
1 mol of 12C atoms = 6.022 ×10²³ atoms = 12g
∴ Mass of 1 atom 12C = 12 /6.022 ×10²³ g = 1.9927× 10^-23 g

31. How many significant figures should be present in the answer of the following calculations?
(i) 0.02856 × 298.15 × 0.112 /0.5785
(ii) 5 × 5.364
(iii) 0.0125 + 0.7864 + 0.0215
Solution :
(i) Least precise term i.e. 0.112 is having 3 significant digits.

∴ There will be 3 significant figures in the calculation.

(ii) 5.364 is having 4 significant figures.

∴ There will be 4 significant figures in the calculation.

(iii) Least number of decimal places in each term is 4.

∴ There will be 4 significant figures in the calculation.

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

Isotope              Isotopic molar mass            Abundance

36Ar                  35.96755 g mol^-1                 0.337%

38Ar                  37.96272 g mol^-1                0.063%

40Ar                  39.9624 g mol^-1                  99.600%
Solution :
Molar mass of Ar =  ∑piAi
= (0.00337 × 35.96755 )+ (0.00063 × 37.96272 )+(0.99600 × 39.9624 ) = 39.948 g mol^-1

33. 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.

Solution :

(i) 1 mol of Ar = 6.022 × 10²³atoms

∴ 52 mol of Ar = 52 × 6.022×10²³atoms = 3.131 × 10²⁵atoms

(ii) 1 atom of He = 4 u of He

4 u of He = 1 Atom of He

∴ 52 u of He = 1/4 × 52 = 13 atoms

(iii) 1 mol of He = 4 g = 6.022 × 10²³atoms

∴ 52 g of He = (6.022 × 10²³/4) × 52 atoms = 7.8286 × 10²⁴atoms

34. 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, and (iii) molecular formula.

Solution :

Amount of carbon in 3.38 g of CO₂ = 12/44 × 3.38 g = 0.9218 g
Amount of hydrogen in 0.690 g H₂O = 2/18 × 0.690 g = 0.0767 g
The compound contains only C and H, therefore total mass of the compound = 0.9218 + 0.0767 = 0.9985 g
% of C in the compound = (0.9218 /0.9985 )×100 = 92.32
% of H in the compound = (0.0767 /0.9985 )×100 = 7.68
(i) Calculation of empirical formula,
Moles of carbon in the compound = 92.32/12 = 7.69
Moles of hydrogen in the compound = 7.68/1 = 7.68
Simplest molar ratio = 7.69 : 7.68 = 1(approx)
∴ Empirical formula CH
(ii) 10.0 L of the gas at STP weigh = 11.6 g
∴ 22.4 L of the gas at STP = 11.6/10.0 × 22.4 = 25.984 = 26 (approx)
∴ Molar mass of gass = 26 g mol-1
(iii) Mass of empirical formula CH = 12+1 = 13

∴ n = Molecular Mass/Empirical Formula = 26/13 = 2 

∴ Molecular Formula = C₂H₂

35. Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction, CaCO₃ (s) + 2HCl (aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)
What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl?
Solution :
1000 mL of 0.75 M HCl have 0.75 mol of HCl = 0.75×36.5 g = 24.375 g
∴ Mass of HCl in 25mL of 0.75 M HCl = 24.375/ 1000 × 25 g = 0.6844 g
From the given chemical equation,
CaCO₃ (s) + 2HCl (aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)
2 mol of HCl i.e. 73 g HCl react completely with 1 mol of CaCO₃ i.e. 100g
∴ 0.6844 g  HCl reacts completely with CaCO₃ = 100/73 × 0.6844 g = 0.938 g

36. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO₂) with aqueous hydrochloric acid according to the reaction
4HCl (aq) + MnO₂(s) → 2H₂O(l) + MnCl2(aq) + Cl₂(g)
How many grams of HCl react with 5.0 g of manganese dioxide?
Solution :
1 mol of MnO₂ = 55+32 g = 87 g
87 g of MnO₂ react with 4 moles of HCl i.e. 4×36.5 g = 146 g of HCl.
∴ 5.0 g of  MnO₂ will react with HCl = 146/87×5.0 g = 8.40 g.

MCQ

1. Which of the following is dependent on temperature?

(a) Molarity (b) Molality (c) Mole fraction (d) Mass percentage

2. If the concentration of glucose (C₆H₁₂O₆) in blood is 0.9 g L molarity of glucose in blood?

a) 5M b) 50M c) 0.005 M d) 0.5 M

3. What is the mass percent of carbon in carbon dioxide?

a) 0.034% b) 27.27% c) 3.4% d) 28.7%

4. The empirical formula and molecular mass of a compound are CH₂O and 180 g respectively. What will be the molecular formula of the compound?

a) C₉H₁₈O₉ b) CH₂O c) C₆H₁₂O₆ d) C₂H₄O₂

5. the total number of ions present in 111 g of CaCl₂ is

(a) One Mole (b) Two Mole (c) Three Mole (d) Four Mole

6. Which one will have maximum numbers of water molecules?

(a) 18 molecules of water (b) 1.8 grams of water

(c) 18 grams of water (d) 18 moles of water

7. There are two chlorides of sulphur S₂Cl₂ and SCl₂. What is the equivalent mass of S in  SCl₂

(a) 64.8 g/mole (b) 32 g/mole (c) 16 g/mole (d) 8 g/mole

8. Which is not a unit of pressure:

(a) Bar (b) N/m² (c) Kg/m² (d) Torr

9. What is the normality of a 1 M solution of H₃PO₄

(a) 0.5 N (b) 1.0 N (c) 2.0 N (d) 3.0 N

10. The significant figures in 3400 are

(a) 2 (b) 5 (c) 6 (d) 4

11. Which of the following contains the same number of carbon atoms as are in 6.0 g of carbon (C – 12)?

(a) 6.0 g Ethane (b) 8.0g Methane (c) 21.0g Propane (d) 28.0 g CO

ANSWER:

1. (a)
2. (c)
3. (b)
4. (c )
5. (c)
6. (d)
7. (c)
8. (c)
9. (d)
10. (d)
11. (b)

ASSERTION AND REASON QUESTIONS

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given.

Choose the correct option out of the choices given below each question.

a) If both Assertion & Reason are true and the reason is the correct explanation of the assertion.

b) If both Assertion & Reason are true but the reason is not the correct explanation of the assertion.

c) If Assertion is a true statement but Reason is false.

d) If both Assertion and Reason are false statements.

1. Assertion (A): The empirical mass of ethene is half of its molecular mass.

Reason (R): The empirical formula represents the simplest whole number ratio of various atoms present in a compound.

Answer : (b)

2.Assertion (A) : One atomic mass unit is defined as one twelfth of the mass of one carbon-12 atom.

Reason (R) : Carbon-12 isotope is the most abundant isotope of carbon and has been chosen as standard.

Answer : (b)

3.Assertion (A) : Significant figures for 0.200 is 3 whereas for 200 it is 1.

Reason (R) : Zero at the end or right of a number are significant provided they are not on the right side of the decimal point.

 Answer: (c)

4. Assertion (A): Combustion of 16 g of methane gives 28 g of water.

Reason (R): In the combustion of methane, hydrogen is one of the products.

 Answer: (d)

5. Assertion: Molarity is number of moles of solute in 1 lit of solution

Reason: Molality does not change with temperature.

Answer: (b)

One Mark questions :

1.Calculate the molecular mass of the of the following

(i)H₂SO₄ (ii)NaOH

Ans: Molar mass of H2SO4= 2+32+4*16=98 amu

Molar mass of NaOH = 23+16+1=40amu

2. What do mean by Mole fraction?

Ans. Mole Fraction is the ratio of number of moles of one component to the total number of moles (solute and solvents) present in the solution. It is expressed as ‘x’.

3. What is the limiting reagent?

Ans. The reactant which gets consumed first or limits the amount of product formed is known as limiting reagent

4. What is the relation between temperature in degrees Celsius and degree Fahrenheit?

Ans.

5. Define one mole?

Ans. One mole is the amount of a substance that contains as many particles as there are atoms in exactly 12 g of the carbon-12.

6. Write the empirical formula of the following:

(a)N₂O₄ (b)C₆H₁₂O₆ (c) H₂O (d) H₂O₂

Ans. (a)NO₂ (b) CH₂O (c) H₂O (d) HO

7. Briefly explain the difference between precision and accuracy.

Ans. Precision refers to the closeness of various measurements for the same quantity. However, accuracy is the agreement of a particular value to the true value of the result.

8. Define the law of multiple proportions. Explain it with one example.

Ans. 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, bear a simple ratio to one another. For example- carbon combines with oxygen to form two compounds CO and CO₂.

Compound  CO   CO₂

Mass of C     12.  12

Mass of O    16    32

Masses of oxygen which combine with a fixed mass of carbon (12g) bear a simple ratio of

16:32 or 1:2.

9. Chlorine has two isotopes of atomic mass units 34.97 and 36.97. The relative abundance of the isotopes is 0.755 and 0.245 respectively. Find the average atomic mass of chlorine.

Ans. Average atomic mass = 34.97 x 0.755 +36.97 x 0.245 = 35.46 u

10. How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?

Ans. Molar mass of Na2CO3= 2 x 23 +12 + 3 x 16 = 106 g / mol

0.50 mol Na₂CO₃and 0.50 x 106 = 53 g

0.50 M Na₂CO₃ means 0.50 mol i.e. 53 g of Na₂CO₃ are present in I L of the solution.

Three Marks questions

1.How many molecules approximately do you expect to be present in a small crystal of sugar which weighs 10 mg?

Answer:

10 mg sugar (C₁₂H₂₂O₁₁) = 0.01 g = 0.01/342 mol

= 2.92 × 10^-5 mole

= 2.92 × 10^-5 × 6.02 × 10²³ molecules

= 1.76 × 10¹⁹ molecules.

2. What do you mean by molarity? Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution.

And. The number of moles of solute dissolved per litre (dm³) of the solution is called molarity

Since molarity (M) =

3. Express the following in the scientific notation with 2 significant figures-

(a) 0.0048 (b) 234,000 (c) 200.0

Ans. (a) 4.8 x 10^-3 (b) 2.3 x 10⁵ (c) 2.0 x 10²

4.Calcium carbonate reacts with aqueous HCl according to the reaction

CaC0₃ (s) + 2HCl (aq) ———->CaCl₂ (aq) +C0₂(g) +H₂O(l).

What mass of CaC0₃ is required to react completely with 25 mL of 0.75 M

Ans. Step 1. To calculate the mass of HCl in 25 mL of 0.75 m HCl

1000 mL of 0.75 M HCl contain HCl = 0.75 mol = 0.75 x 36.5 g = 24.375 g

Step 2. To calculate mass of CaC0₃ reacting completely with 0.9125 g of HCl

CaC0₃ (s) + 2HCl (aq) ———->CaCl₂ (aq) +C0₂(g) +H₂O(l).

2 mol of HCl, i.e., 2 x 36.5 g = 73 g HCl react completely with CaC0₃ = 1 mol = 100 g

5. Chlorine is prepared in the laboratory by treating manganese dioxide (Mn02) with aqueous hydrochloric acid according to the reaction.

4 HCl (aq) + Mn0₂ (s) ———–> 2H2O (l) + MnCl₂(aq) +Cl₂(g)

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

(Atomic mass of Mn = 55 u)

Answer: 1 mole of Mn0₂, i.e., 55 + 32 = 87 g

Mn0₂ react with 4 moles of HCl, i.e., 4 x 36.5 g = 146 g of HCl.

Case based questions

1 . The mass of one mole of a substance in grams is called its molar mass. the molar mass in grams is numerically equal to atomic molecular/formula mass in u. An empirical formula represents the simplest whole number ratio of various atoms present in a compound, whereas, the molecular formula shows the exact number of different types of atoms present in a molecule of a compound. If the mass per cent of various elements present in a compound is known, its empirical formula can be determined. Molecular formula can further be obtained if the molar mass is known. Many a time, reactions are carried out with the Amounts of reactants that are different than The amounts as required by a balanced chemical reaction. In such situations, one Reactant is in more amount than the amount required by balanced chemical reaction. The reactant which is present in the least amount Many a time, reactions are carried out with the amounts of reactants that are different than the amounts as required by a balanced chemical reaction. In such situations, one reactant is in more amount than the amount required by balanced chemical reaction. The reactant which is present in the least amount gets consumed after sometime and after that further reaction does not take place whatever be the amount of the other reactant. Hence, the reactant, which gets consumed first, limits the amount of product formed and is, therefore, called the limiting reagent.

1) One atomic mass unit (amu) is defined as a mass exactly equal to one-twelfth of the mass of one …atom.

(a) Hydrogen – 1 (b) Carbon – 12 (c) Oxygen -12 (d) Chlorine – 35

Ans – b) carbon – 12

2) The mass of one mole of a substance in grams is called its..

(a) Atomic mass (b) Molecular Weight (c) Molecular mass (d) molar mass.

Ans – d) molar mass

3) … is the sum of atomic masses of the elements present in a molecule.

(a) Atomic mass (b) Molecular Weight (c) Molecular mass (d) molar mass.

Ans – c) Molecular mass

4) One mole contains exactly …elementary entities.

(a) 6. 02214076 × 10²¹ (b) 6. 02214076 × 10²² (c) 6.02214076 × 10²³ (d) 6.02214076 × 10²⁴

Ans – c) 6.02214076 × 10²³

Five Marks questions

1. What is the difference between empirical and molecular formula? A compound contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas?

Ans. An empirical formula represents the simplest whole number ration of various atoms present in a compound whereas the molecular formula shows the exact number of different types of atoms present in a molecule of a compound

The empirical formula of the above compound is CH₂Cl.

empirical formula mass is 12 + (1×2) + 35.5 = 49.5

n= molecular mass/ empirical formula mass =98.96/49.5 = 2

Hence molecular formula is C₂H₄Cl₂

2. Dinitrogen and dihydrogen react with each other to produce ammonia according to the

following chemical equation:(i) N₂ (g) + 3H₂(g) —–> 2NH₃ (g)

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

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

Answer:

3. 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) the molar mass of the gas, and

 (iii) molecular formula.

Ans: We know that

44g of CO₂ contains 12 g of carbon.

3.38 g of CO₂ will contain carbon

18 g of water contains 2 g of hydrogen.

0.690 g of water will contain hydrogen =

Since carbon and hydrogen are the only constituents of the compound, the total mass of

the compound is:= 0.9217 g + 0.0767 g= 0.9984 g

Percent of C in the compound =

Percent of H in the compound =

Moles of carbon in the compound =

Moles of hydrogen in the compound =

Ratio of carbon to hydrogen in the compound = 7.69: 7.68= 1: 1

Hence, the empirical formula of the gas is CH.

(ii) Given,

Weight of 10.0L of the gas (at S.T.P) = 11.6 g

Weight of 22.4 L of gas at STP = = 25.984 g=26g

Hence, the molar mass of the gas is 26 g.

Empirical formula mass of CH = 12 + 1 = 13 g

n = = 2

Molecular formula of gas = (CH)n= C₂H₂

HOTS (Higher Order Thinking Skills

1. A compound made up of two elements A and B has A= 70 %, B = 30 %. Their relative number of moles in the compound are 1.25 and 1.88. calculate

a. Atomic masses of the elements A and B

b. Molecular formula of the compound, if its molecular mass is found to be 160

Ans: No. of moles = , Atomic mass =

Atomic mass of A = =

Atomic mass of B = =

Empirical formula

Element Relative

no. of

moles

Simplest molar ratio Simplest whole

no.

molar ratio

A 1.25 1.25/1.25 = 1 2

B 1.88 1.88/1.25 = 1.5 3

Empirical formula = A2B3

Calculation of molecular formula-

Empirical formula mass = 2 x 56 + 3x 16 = 160

N= =

Molecular formula = A2B3