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Question (1):
What is a symbol? What information does it convey? |
Answer:
A short hand representation of an element is called symbol. It represents the following:
 Name of the element
One atom of the element
One mole of atoms. It represents 6.023 x 1023 atoms of the element.
A definite mass of the element
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Question (2):
What is the difference between symbol of an element and formula of an element? |
| Answer:
Symbol of an element represents the name of the element. It also represents one atom of the element.
Example: H represents hydrogen and C represents carbon
A formula of an element represents the number of atoms in the molecule of the compound. One molecule of hydrogen element contains two atoms of hydrogen; therefore the formula of hydrogen is H2.
2H represents two separate atoms of hydrogen, whereas H2 represents 1 molecule of hydrogen similarly the molecular formula of oxygen element and chlorine element is O2 and Cl2.
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Question (3):
Define atomic mass of an element. |
| Answer:
The atomic mass of an element is the relative mass of its atom as compared to the mass of C - 12 atom taken as 12 units. |
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Question (4):
State the laws of chemical combination. |
Answer:
Laws of chemical combinations are:
i) Law of Conservation of Mass: The law states that during any physical or chemical change, the total mass of the product remains equal to the total mass of the reactants.
ii) Law of Constant Composition: The law states that a chemical compound always contains same elements combined together in the same proportion by mass.
iii) Law of Multiple Proportions: The law states that when two elements combine with each other to form two or more compounds, the masses of one of the elements, which combine with fixed mass of the other, bear a simple whole number ratio to one another. |
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Question (5):
In an experiment it was found that litharge, red oxide of lead and lead peroxide contained 92.83%, 90.6% and 86.6% of lead respectively. Show that these figures are in agreement with the Law of Multiple Proportions. |
Answer:
In litharge, the amount of lead = 92.83%
The amount of oxygen = 100 - 92.83 = 7.17%
7.17g of O2 combines with 92.83g of lead
In red oxide of lead, the amount of Pb = 90.6%
The amount of oxygen = 100 - 90.6 = 9.4 %
9.4 g of oxygen combines with 90.6g of lead
In lead peroxide, the amount of lead = 86.6%
The amount of oxygen = 100 - 86.6 = 13.4%
13.4g of oxygen combines with 86.6g of lead
Thus, the different weights of lead combining with fixed weight (1g) of oxygen are in the ratio 12.947: 9.638: 6.462 = 2:1.5: 1 or 4: 3: 2
This is in line with the law of multiple proportions.
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Question (6):
What do you understand by atomicity of an element? Give an example of a polyatomic molecule. |
| Answer:
The number of atoms that constitute one molecule of an element is called its atomicity.
Sulphur is a polyatomic molecule: S8 - 8 is the atomicity.
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Question (7):
What do the following stand for?
(i) P4 and 4P (ii) O2 and 2O
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Answer:
P4 - 1 molecule of phosphorus
4P - 4 atoms of phosphorus
O2 - 1 molecule of oxygen
2O - 2 atoms of oxygen
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Question (8):
How many gram atoms are present in 69 grams of sodium? |
Answer:
The number of gram atoms present in 69 grams of sodium is 3.
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Question (9):
The mass of a single atom of an element Z is 2.65x10-23g. What is its gram atomic mass? |
Answer:
1 atom of element Z has mass = 2.65 x 10-23g
Hence, 6.023 x 1023 atoms of element Z have mass
= 2.65 x 10-23 x 6.023 x 1023
= 15.69 g
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Question (10):
What is gram molecular mass? |
| Answer:
The amount of a substance whose mass in grams is numerically equal to its molecular mass is called gram molecular mass. |
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Question (11):
Calculate the molar mass of HNO3. [N = 14, O = 16, H = 1] |
Answer:
Molar mass of HNO3.
H = 1 x 1 = 01
N = 14 x 1 = 14
O = 16 x 3 = 48
Total mass = 63 grams
Molar mass of HNO3= 63 grams
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Question (12):
Calculate the formula mass of CaCl2. [Ca = 40, Cl = 35.5] |
Answer:
1(Ca) + 2(Cl)
40 + 2x(35.5) = 111 amu
The formula mass of CaCl2 is 111 amu.
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Question (13):
A certain non-metal X forms two oxides I and II. The mass percentage of oxygen in oxide I (X4O6) is 43.7, which is same as that of X in oxide II. Find the formula of the second oxide. |
Answer:
Now 43.7 parts of oxygen in I corresponds to = 6 oxygen atoms
Also 56.3 parts of X in I correspond to = 4 X atom
Now the atomic ratio X : O in the second
The formula of the second oxide is X2O5.
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Question (14):
(i) Calculate the mass of 0.2 moles of water (O=16, H=1).
(ii) What is the volume of 7.1 g of chlorine (Cl=35.5) at S.T.P.
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Answer:
(i) Gram Molecular Weight of H2O = 2 x 1 + 16 = 18 g
1 mole of water weighs 18 g
(ii) Gram Molecular Weight of Cl2 (one mole)= 35.5 x 2 = 71 g.
71 g of Cl2 at S.T.P occupies 22.4 litres
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Question (15):
The reaction between aluminium carbide and water takes place according to the following equation:
Calculate the volume of CH4 released from 14.4 g of Al4C3 by excess water at S.T.P. (C = 12, Al = 27)
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Answer:
Molecular weight of Al4C3 is (27 x 4) + (12 x 3) = 144
144 g of Al4C3 produces 3 x 22.4 litres of CH4 at S.T.P
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Question (16):
A compound of sodium, sulphur and oxygen has the following percentage composition. Na=29.11%, S=40.51%, O=30.38%. Find its empirical formula (O=16, Na=23, S=32). |
Answer:
Empirical formula is NaSO1.5 or to its nearest whole number i.e., the formula is Na2S2O3.
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Question (17):
Solid ammonium dichromate with relative molecular mass of 252 g decomposes according to the equation.
(i) What volume of nitrogen at S.T.P will be evolved when 63 g of (NH4)2Cr2O7 is decomposed?
(ii) If 63 g of (NH4)2Cr2O7 is heated above 1000C, what will be the loss of mass? (H=1, N=14, O=16, Cr=52).
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Answer:
252 g of (NH4)2Cr2O7 gives one mole or 22.4 litres of N2 at S.T.P as per the given equation.
(ii) At temperatures above 1000C water is in the form of steam.
Products as vapours are N2 and H2O.
The transformation of solids and liquids into gaseous substances results in loss of mass.
Total weight of gaseous products = {(2 x 14) + 4 (2 x 1) + 16}
= 28 + 72 = 100 g
Heating 252 g of (NH4)2Cr2O7 causes 100 g loss of mass.
The loss of mass is 25 g
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Question (18):
How many litres of ammonia are present in 3.4 kg of it? (N = 14, H = 1) |
Answer:
Gram molecular weight of NH3 = 14 + (1 x 3) = 17 g.
17 g of NH3 = 22.4 litres
= 
= 4480 litres.
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Question (19):
About 640 mL of carbon monoxide is mixed with 800 mL of oxygen and ignited in an enclosed vessel. Calculate the total volume of gases after the burning is completed. All volumes are measured at S.T.P. |
Answer:
The chemical reaction actually taking place is:
Volume of O2 used = 320 mL.
Volume of O2 left = 800 - 320 = 480 mL.
Volume of CO2 formed = 640 mL.
Therefore the total volume of gases after burning is 480 + 640 = 1120 mL.
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Question (20):
Calculate the number of moles of ammonium sulphate present in 15.84 kg of it. (H=1, N=14, O=16, S=32) |
Answer:
Molecular weight of (NH4)2SO4
= (2 x 14) + (2 x 4) + 32 + (16 x 4)
= 132 a.m.u.
132 g of (NH4)2SO4 = 1 mole
The number of moles of ammonium sulphate present in 15.84 kg is 120 moles.
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Question (21):
What is the mass of 0.2 mole of lead nitrate? (N=14, O=16, Pb=207). |
Answer:
Gram molecular weight of Pb(NO3)2 = 207 + (2 x 14) + 2(16 x 3)
= 207 + 28 + 96
= 331
1 mole of Pb(NO3)2 is 331 g
Therefore 0.2 mole of Pb(NO3)2 is 331 x 0.2 = 66.2 g |
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Question (22):
Find the total percentage of oxygen in magnesium nitrate crystals i.e.,
Mg(NO3)2.6H2O (Atomic weight: H=1, N=14, O=16, Mg=24).
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Answer:
Molecular weight of Mg(NO3)2.6H2O
= 24 + 2(14 + 16 x 3) + 6(2 x 1 + 16)
= 24 + 124 + 108 = 256 a.m.u
Atomic mass of oxygen in Mg(NO3)2.6H2O is,
= 2 (16 x 3) + 6 (16) = 96 + 96 = 192
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Question (23):
A compound has the following percentage composition H=2.04%, S=32.65%, O=65.31%. Relative molecular mass of the compound = 98. Calculate its molecular formula (H = 1, S = 32, O = 16). |
Answer:
Empirical formula is H2SO4
Empirical formula mass = (2 x 1) + 32 + (16 x 4) = 98.
Relative molecular mass = 98
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Question (24):
Calculate the amount of nitrogen supplied to soil by 1 quintal (100 kg) of ammonium nitrate (N=14, H=1, O=16). |
Answer:
Molecular weight of NH4NO3 = 14 + (4 x 1) + 14 + (16 x 3)
= 80 g
Molecular weight of N in the above formula = 14 x 2 = 28
80 units of NH4NO3 yield 28 units of Nitrogen.
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Question (25):
Identify diatomic molecules from the following:
(i) HCl (ii) P4 (iii) He (iv) O3 (v) H2S (vi) CO
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| Answer:
HCl, CO are diatomic. |
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