The matter around us is made up of small particles called atoms. The substances made up of the same kind of atoms are called Elements. The substances made from the same kind of groups of atoms are called compounds. Both the elements and compounds are pure substances.
| Definition: An Element is the simplest form of a kind of matter having definite physical and chemical properties. The elements contain only one kind of atom; while the compounds contain atoms of more than one kind. |
Elements and Compounds Chemistry Questions with Solutions
Q1: 500 mL of a 5 M sample is diluted to 1500 mL. What will be the molarity of the new solution formed?
- 1.59 M
- 1.66 M
- 0.017 M
- 1.5 M
Answer: (2)
Explanation: From the molarity formula: M1V1 = M2V2
Here M1 = 5 M, V1 = 500 mL, M2 = ? and V2 = 1500 mL
Hence, 5 M x 500 mL = M2 x 1500 mL
M2 = 5/3 = 1.66 M
Q2. At STP, one mole of oxygen gas is equal to _____.
- 16 g of oxygen
- 32 atoms of oxygen
- 6.022 x 1023 atoms of oxygen
- 6.022 x 1023 molecules of oxygen
Answer: (4)
Explanation: 1 mole of oxygen gas at STP = 6.022 x 1023 molecules of oxygen = 32 g of oxygen
Q3. Five grams of each of the following gases are taken at 87 oC and 750 mm pressure. Predict which of these gases will have the least volume.
- HF
- HCl
- HBr
- HI
Answer: (4)
Explanation: From the mole concept: Number of moles = given mass/ molecular formula mass
Hence, number of moles in 5 g of HF = 5/ 20 mol
Number of moles in 5 g of HCl = 5/36.5
Number of moles in 5 g of HBr = 5/81
Number of moles in 5 g of HI = 5/121
From the calculation of the number of moles, it is clear that 5 g of HI has the least number of moles among other gases. Hence, the least number of moles correspond to least volume.
Q4. 8 g of O2 has the same number of molecules as
- 7 g of CO
- 11 g of CO2
- 16 g of SO2
- All of the above
Answer: (4)
Explanation: As Number of moles = given mass/ molecular formula mass
Number of moles in 8 g of O2 = 8/32 = 0.25 mol
Number of moles in 11 g of CO2 = 11/44 = 0.25 mol
Number of moles in 7 g of CO = 7/28 = 0.25 mol
Number of moles in 16 g of SO2 = 16/64 = 0.25 mol
Q5. What are Polymorphs? Give an example.
Answer: The compound that exists in different crystalline forms is said to exhibit the property of Polymorphism. The different crystalline forms of the compound are called Polymorphs.
For example: ZnS has two polymorphs namely Zinc Blende and Wurtzite. Both Zinc Blende and Wurtzite have the chemical formula ZnS but differ in their structures. Zinc Blende has a cubic while wurtzite has a hexagonal structure.
Q6. What is the difference between the molecular mass and the gram molecular mass?
Answer: The molecular mass of a molecule is the actual mass of the molecule expressed in grams. However, the gram molecular mass is the mass of Avagadro’s number of molecules (6.022 x 1023 molecules) in grams.
Q7. The two isotopes of chlorine account for its molecular mass i.e. 35.45 amu. Calculate the percentage of both the naturally occurring isotopes of chlorine.
Answer: Chlorine has two isotopes: 35Cl and 37Cl.
Let us assume that the percentage of 35Cl in naturally occurring chlorine is 𝒙% and that of 37Cl is (100-𝒙)%.
Hence, Average atomic mass = (%35Cl x mass of 35Cl) + (%37Cl x mass of 37Cl) / 100
Average atomic mass = (𝒙 x 35) + {(100 – 𝒙)} / 100 = 35.45 (given average atomic mass)
So, 𝒙 = 77.5 % and (100 – 𝒙) = 22.5%
Hence, 35Cl = 77.5% and 37Cl = 22.5%.
Q8. Calculate the molecular mass of water that contains 50% heavy water (D2O).–
Answer: As is given in the question, the sample of water contains 0.5 mole of H2O and 0.5 mole of D2O.
Mass of 1 mole of H2O = 18 g
Mass of 0.5 moles of H2O = 0.5 x 18 g = 9 g
Similarly, Mass of 1 mole of D2O = 20 g
Mass of 0.5 moles of D2O = 0.5 x 20 g = 10 g
Since the mass of 1 mole of a substance is called its molar mass. The molar mass of the given sample of water = 10 g + 9 g = 19 g mol-1.
Q9. The mass of the black dot used at the end of a sentence is considered to be 1 Attogram. If the black dot is made of carbon, calculate the number of moles of carbon present in the black dot.
Answer: Given: mass of carbon in the black dot = 1 Attogram = 10-18 g
Now, 12 g carbon contains atoms = 6.022 x 1023 atoms
10-18 g of carbon contains atoms = ( 6.022 x 1023 atoms / 12 g) x 10-18 g = 5.02 x 104 atoms
So, the black dot contains 5.02 x 104 atoms of carbon.
Q10. What will be the minimum molecular mass of Insulin that contains 3.4% S?
Answer: The minimum molecular mass of Insulin will be the mass that contains the mass of at least 1 atom of S. The atomic mass of S = 32 amu
Now that 3.4 amu of S is present in Insulin = 100 amu
32 amu os S will be present in Insulin = (100/3.4) x 32 amu = 941.2 amu
Hence, the minimum molecular mass of Insulin that contains 3.4% S is 941.2 amu.
Q11. Find the normality of the acid solution when 10 mL of HCl solution is treated with excess of AgNO3 to give 0.1435 g of AgCl.
Answer: 143.5 g AgCl contains Cl = 35.5 g
0.1435 g of AgCl contains Cl = (35.5 / 143.5) x 0.1435 g = 0.0355 g
Now, 35.5 g Cl is contained in HCl = 36.5 g
0.0355 g Cl is contained in HCl = (36.5 / 35.5) x 0.0355 g = 0.0365 g
Now that, Number of gram equivalents = mass of solute (g) / Equivalent mass of solute
Hence, Gram equivalents of 0.0365 g HCl = (0.0365 / 36.5) g eq. = 0.001 g eq.
10 mL of HCl contains 0.001 g eq.
1000 mL of HCL will contain g eq. = (0.001/10) x 1000 = 0.1 g eq.
As we know, the gram equivalents present in 1 L of the solution is called Normality.
Hence, the Normality of the given acidic solution is 0.1 N.
Q12. Calculate the atomicity of a molecule of H2SO4.
Answer: The total number of atoms present in a molecule accounts for its atomicity.
Hence, the atomicity of H2SO4 = 2 + 1 + 4 = 7.
Q13. Count the number of significant figures in 1.050 x 104.
Answer: In the scientific notation (N x 10n), the number of digits present in N are counted in significant figures. Hence, 1.050 x 104 has 4 significant figures.
Q14. What is the S.I. unit of Density?
Answer: The density formula is: Density = Mass / Volume
The S.I. unit of mass is kg, while that of volume is m3. Hence, the S.I. unit of Density is kg m-3.
Q15. Match the following:
| (i) | 88 g of CO2 | (a.) | 0.25 mol |
|---|---|---|---|
| (ii) | 96 g of O2 | (b.) | 6.022 x 1023 molecules |
| (iii) | 1 mole of any gas | (c.) | 2 mol |
| (iv) | 5.6 L of CO2 at STP | (d.) | 3 mol |
Answer: (i)-(c.), (ii)-(d.), (iii)-(b.), (iv)-(a.)
Explanation: (i) Mass of 1 mole of CO2 = 44 g
Hence, 88 g of CO2 corresponds to 2 moles.
(ii) 32 g of O2 = 1 mol
96 g of O2 = (1 / 32) x 96 mol = 3 mol
(iii) 1 mole of any gas = 6.022 x 1023 molecules
(iv) 22.4 L of CO2 at STP = 1 mol
5.6 L of CO2 at STP = (1 / 22.4) x 5.6 mol = 0.25 mol
Practise Questions on Elements and Compounds
Q1. The Empirical formula of hydrogen peroxide is______.
Q2. Calculate the mass percent of carbon in carbon dioxide.
Q3. What is the difference between 2.5 x 103 g and 2.50 x 103 g?
Q4. Why is molality preferred over molarity in expressing the concentration of a solution?
Q5. How much of Cu can be obtained from 100 g of copper sulphate (CuSO4)?
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