Some Basic Concepts of Chemistry Most Important Questions

Q1. A compound prepared by any method contains the same elements in the fixed ratio by mass. The given statement is known as:

  • 1) Law of multiple Proportions
  • 2) Law of conservation of mass
  • 3) Law of reciprocal Proportions
  • 4) Law of definite Proportions

Solution

According to the law of definite proportions, a given compound always contains exactly the same proportion of elements by mass.

Q2. The empirical formula of an organic gas containing carbon and hydrogen is CH2. The mass of 1 litre of this organic gas is exactly equal to that of 1 litre of N2. Therefore, the molecular formula of organic compound is:

  • 1) C6H12
  • 2) C4H8
  • 3) C2H4
  • 4) C3H6

Solution

Molecular mass of the gas = Molecular mass of N2 = 28 g Molecular formula mass = n X (Empirical formula mass) Empirical formula mass of the gas= 12 + 2 = 14 g

begin mathsize 12px style Therefore comma space straight n space equals space 28 over 14 space equals space 2 end style 

Hence the molecular formula is (CH2)2 or C2H4.

Q3. What is the simplest formula of the compound which has the following percentage composition? Carbon 80% and Hydrogen 20%. If the molecular mass is 30, calculate its molecular formula.

Solution

Calculation of empirical formula: Element Percentage Atomic mass Relative number of moles Simple ratio Simplest whole number ratio C     H 80     20 12     1    1     3   Therefore Empirical formula is CH3  

Calculation of molecular formula:  Empirical formula mass = 12 × 1 + 1 × 3 = 15

 begin mathsize 11px style straight n space equals space fraction numerator Molecular space mass over denominator Empirical space formula space mass end fraction space equals space 30 over 15 space equals space 2 end style 

Molecular formula = Empirical formula × 2 = CH× 2 = C2H6

Q4. A statement which is not a part of Dalton’s atomic theory is:

  • 1) Matter is made of atoms.
  • 2) Atoms are composed of sub-atomic particles called electron, proton and neutron.
  • 3) All atoms of a given element are identical.
  • 4) Atoms of different elements have different mass and different properties.

Solution

According to the Dalton’s atomic theory, atoms are indivisible i.e. they are not composed of any sub-atomic particles.

Q5. 18 g of glucose (C6H12O6) is present in 1000 g of an aqueous solution of glucose. The molality of this solution is:

  • 1) 0.5m
  • 2) 1m
  • 3) 1.1m
  • 4) 0.1m

Solution

18 g of glucose (C6H12O6) is present in 1000 g of an aqueous solution of glucose. The molality of this solution is 0.1 M. Solution: Given mass of Glucose = 18 g Mass of Solvent = 1000 g Molar mass of glucose = 180 g

 begin mathsize 11px style Molality space equals space fraction numerator Given space Mass space of space Solute over denominator Molar space Mass space of space Solute space cross times space Molar space Mass space of space Solvent space in space Kg end fraction
Molality space equals space fraction numerator 18 space cross times space 1000 over denominator 180 space cross times space 1000 end fraction
space space space space space space space space space space space space space space space equals space 0.1 space molal end style

Q6. Classify the following substances as elements, compounds and mixtures. In case of mixtures clearly indicate whether the mixture is homogenous or heterogeneous and also explain how would you separate them?

  • a) Iron
  • b) Salt and water
  • c) Glass powder, iron filings and sugar
  • d) Distilled water

Solution

a) Iron is an element

b) Salt and water are a homogenous mixture which can be separated by distillation, in this process salt remains as residue.

c) Glass powder, iron filing and sugar are a heterogeneous mixture. To separate the constituents dissolve the mixture in water, sugar dissolves. Now filter to separate undissolved glass powder and iron filings. Glass powder and iron filings can be separated with the help of a magnet. Sugar can be recovered from solution by evaporation.

d) Distilled water is a compound.

Q7. Gastric juice contains 3.0 g of HCl per litre. If a person produces about 2.5 litres of gastric juice per day, How many antacid tablets each containing 400 mg of Al(OH)3 are needed to neutralize all the HCl produced in one day.

Solution

Al(OH)3  +  3HCl  →  AlCl3  +  2H2O   78 g                    109.5 g   begin mathsize 11px style Amount space of space HCl space produced space equals space 3.0 space cross times 2.5 space equals space 7.5
Amount space of space Al left parenthesis OH right parenthesis subscript 3 space space required space equals space fraction numerator 78 over denominator 109.5 end fraction space cross times space 7.5 space equals space 5.342 space straight g space equals space 5342 space mg
Therefore space number space of space Al left parenthesis OH right parenthesis subscript 3 space tablets space required space equals space 5342 over 400 space equals space 13.35 space equals space 14.0 space tablets end style

Q8. 10 ml of HCl solution produced 0.1435 g of AgCl when treated with excess of silver nitrate solution. What is the molarity of acid solution? (Atomic mass of Ag = 108).

Solution

The chemical equation of the reaction is:      HCl  +  AgNO3   →    AgCl  +  HNO3  36.5 g                     143.5g 

begin mathsize 11px style Amount space of space HCl space mg space required space to space produce space 0.1435 space straight g space AgCl space equals space fraction numerator 36.5 over denominator 143.5 end fraction space cross times space 0.1435 space equals space 0.0365 space straight g
Thus space 10.0 space ml space HCl space contains space HCl space equals space 0.0365 space straight g space equals space fraction numerator 0.0365 over denominator 36.5 end fraction space equals space 10 to the power of negative 3 end exponent space moles
Hence comma space molarity space of space HCl space equals space 10 to the power of negative 3 end exponent over 10 space cross times space 1000 space equals space 0.1 space straight M end style  

Q9. Calculate the number of moles of carbon dioxide which contain 8g of oxygen.

Solution

begin mathsize 12px style In space CO subscript 2 comma space 32 space straight g space of space oxygen space is space present space in space 1 space mole space of space CO subscript 2. space
Therefore comma space 8 space straight g space of space oxygen space is space present space in space equals space fraction numerator 1 space cross times space 8 over denominator 32 end fraction space equals space 0.25 space mol space of space CO subscript 2 end style

Q10. A non-metal X forms two oxides I and II. The mass percentage of oxygen in I (X4O6) is 43.7% which is same as that of X in the 2nd oxide. What is the formula of 2nd oxide?

Solution

Oxygen  X  I oxide 43.7  56.3 II oxide  56.3   43.7   From the data given above we can say that 43.7 parts of Oxygen corresponds to = 6 oxygen atoms

  begin mathsize 11px style therefore space 56.3 space parts space of space Oxygen space in space II space correspond space to space equals fraction numerator 6 space cross times space 56.3 over denominator 43.7 end fraction space equals space 7.730 space atoms
end style 

Also 56.3 parts of X in I correspond to = 4 X atom 

begin mathsize 11px style therefore space 43.7 space parts space of space straight X space in space II space will space correspond space to space equals fraction numerator 4 space cross times space 43.7 over denominator 56.3 end fraction space equals space 3.1 space atoms end stylebegin mathsize 11px style Now comma space the space ratio space of space straight X space colon space straight O space in space second space oxide space equals space fraction numerator 3.1 over denominator 3.1 end fraction space colon space fraction numerator 7.73 over denominator 3.1 end fraction space or space 1 space colon space 2.5 space or space 2 space colon space 5 end style 

Hence, the formula of second oxide is X2O5

Q11. Normality is defined as:

  • 1) Amount of solute present in 100g of solution
  • 2) Number of moles of solute per litre of solution
  • 3) Number of gram equivalents of solute per litre of solution
  • 4) Number of moles of solute per 1000g of solvent

Solution

Normality of a solution is the number of gram equivalents of solute per litre of solution.

Q12. The number of significant figures in 0.04 are:

  • 1) 3
  • 2) 1
  • 3) 2
  • 4) 4

Solution

Zeros before the decimal point are not significant.  

Q13. Carbon forms two gaseous oxides. One of these oxides contains 42.8% carbon while the other contains 27.27% carbon. Explain which law of chemical combination this data illustrates. State this law.

Solution

The ratio of oxygen in the two compounds is 1:2 so this compound follows law of multiple proportions. According to this law when two elements combine with each other to form two or more than two compounds, the masses of one of the elements which combine with fixed mass of the other, bear a simple whole number ratio to the other.

Q14. If 28.036 is rounded off to 4 significant figures, then the result will be:

  • 1) 28.03
  • 2) 28.036
  • 3) 28.0
  • 4) 28.04

Solution

  28.04

Q15. Write a short note on formula unit mass.

Solution

Formula Mass Mass of a molecule of an ionic compound. It is  the sum of atomic masses of all the elements present in one formula unit of a compound. It is used for measuring the molecular mass of entities which do not exist in the solid form.

Substances such as sodium chloride do not contain discrete molecules as their constituent units. In such compounds, positive (sodium) and negative (chloride) entities are arranged in a three-dimensional structure.

In sodium chloride, one Na+ is surrounded by six Cl and vice versa. Hence, in such cases, the formula mass is used to calculate the mass instead of the molecular mass. Just like the molecular mass, it is calculated by adding the atomic masses of the atoms present in one formula unit.

Thus, formula mass of sodium chloride = atomic mass of sodium + atomic mass of chlorine       

 = 23.0 u + 35.5 u = 58.5 u

Q16. How many moles of electrons weight one kilogram?

(Mass of electron = 9.1 Χ 10-31kg, avogadro number = 6.023 Χ 10 23)

  • 1) begin mathsize 12px style fraction numerator 1 space cross times space 10 to the power of 31 over denominator 9.108 end fraction end style
  • 2) begin mathsize 12px style fraction numerator 6.023 space over denominator 9.108 end fraction space cross times space 10 to the power of 54 end style
  • 3) begin mathsize 12px style fraction numerator 1 over denominator 9.108 space cross times space 6.023 end fraction space cross times 10 to the power of 8 end style
  • 4) 6.023 Χ 1023

Solution

Given: Mass of one electron = 9.108 x 10-31 kg 

begin mathsize 12px style Therefore comma space 1 space kg space of space electrons space equals fraction numerator 1 over denominator 9.108 space cross times space 6.023 end fraction space cross times 10 to the power of 8
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 10 to the power of 31 over denominator 9.108 space cross times space 6.023 space cross times space 10 to the power of 23 end fraction space mole
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 10 to the power of 31 space to the power of negative space 23 end exponent over denominator 9.108 space cross times space 6.023 end fraction space mole
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 10 to the power of 8 over denominator 9.108 space cross times space 6.023 end fraction space mole end style

Q17. Number of moles of hydroxide ions in 0.3L of 0.005M solution of Ba (OH)2 is:

  • 1) 0.0075
  • 2) 0.003
  • 3) 0.0015
  • 4) 0.0050

Solution

No of moles = Molarity x volume in litres = 0.005 × 0.3 = 0.0015 moles of Ba(OH)2 1mole of  Ba(OH)2 yields 2 moles of hydroxide ions

Q18. What are significant figures? Give the number of significant figures in each of the following?

  • a) 2.56 Χ 103 
  • b) 165
  • c) 2.05
  • d) 5000
  • e) 0.00256

Solution

The number of significant figures in a measurement is the number of figures that are known with certainty plus one that is uncertain, beginning with one non-zero digit.

a) Number of significant figures in 2.56 Χ 103= 3

b) Number of significant figures in 165 = 3 (all non-zero digits are significant).

c) Number of significant figures in 2.05 = 3 ( zero’s between non zero digits are significant)

d) Number of significant figures in 5000 = 1 (if a number ends in zero’s that are not to the right of a decimal, the zeros may or may not be significant.)

e) Number of significant figures in 0.00256 = 3 (zero’s to the left of the first non-zero digit in the number are non significant.)

Q19. A chloride of phosphorus contains 22.57% P. Phosphine contains 8.82% hydrogen and hydrogen chloride gas contain 97.26% chlorine. Show that the data illustrates law of reciprocal proportions:

Solution

i) In phosphorus chloride, P = 22.57 %,        Cl = 77.43%  

ii) In phosphine,   P = 91.18%, H = 8.82% 

begin mathsize 11px style 22.57 space parts space by space mass space of space straight P space combine space with space straight H space equals space fraction numerator 8.82 over denominator 91.18 end fraction space cross times space 22.57 space equals space 2.18 end style 

Thus, ratio of the masses of Cl and H which combine with fixed mass (22.57 parts) of phosphorus separately is, 77.43: 2.18   or   35.5: 1  

iii) In hydrogen chloride   Cl = 97.26%         H = 2.77%   Thus, ratio of masses of Cl and H when they combine with each other is   97.26: 2.77        or    35.5: 1  

iv) Ratio 1: Ratio 2 = begin mathsize 11px style Ratio space 1 colon space Ratio space 2 space equals fraction numerator 35.5 over denominator 1 end fraction space cross times space fraction numerator 1 over denominator 35.5 end fraction space equals space 1 space colon space 1 end style This shows a simple whole number ratio.

Q20. 1.375 g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper that remained was 1.098g. In another experiment 1.179 g of copper was dissolved in nitric acid and the resulting copper nitrate was converted into cupric oxide by ignition .the weight of cupric oxide formed was 1.476 g. which law of chemical combinations does this data state?

Solution

In first experiment: Copper oxide = 1.375 g Copper left = 1.098 g Therefore oxygen present = 1.375 – 1.098g = 0.277 g

 begin mathsize 11px style Hence space % space of space oxygen space present space in space CuO space equals fraction numerator 0.277 space cross times space 100 over denominator 1.375 end fraction space equals space 20.14 end style 

Second experiment: Copper taken = 1.179 g Copper oxide formed = 1.476 g Therefore oxygen present = 1.476-1.179g = 0.297 g 

begin mathsize 11px style Hence space % space of space oxygen space of space CuO space equals fraction numerator 0.297 space cross times space 100 over denominator 1.476 end fraction space equals space 20.12 end style 

Since percentage of oxygen is the same in both the above cases, so the law of constant composition is illustrated.

Q21. What is the mass (mg) of a zinc block whose dimensions are 2.0″ Χ  3.0″ Χ 5.0″ and whose density is 2.5g /cm3? (Given 1.0″= 2.54 cm.)

Solution

Total volume of a zinc block = 2.0 Χ 3.0 Χ 5.0 = 76.20 Hence mass in grams = density Χ volume = 2.5 Χ 76.20 = 190.500 g = 1.905 Χ 102 g

Q22. If 0.5 mole of BaCl2 is mixed with 0.2 mole of Na3PO4, then the maximum number of moles of Ba3(PO4)that can be formed will be:

  • 1) 0.5
  • 2) 0.3
  • 3) 0.1
  • 4) 0.7

Solution

3BaCl2  +  2Na3PO4  → Ba3(PO4)+ 6NaCl From the molar ratio, we get, BaCl2  :  2Na3PO4  :  Ba3(PO4)2  : 6NaCl   3: 2:1:6 Also, limiting reactant is Na3PO4 Therefore, 0.2 mol Na3PO4 will give Ba3(PO4)2 = 0.5 Χ 0.2 = 0.1 mol.

Q23. According to Dalton’s atomic theory chemical reactions involve:

  • 1) reorganisation of nuclei
  • 2) destruction of atoms
  • 3) reorganisation of atoms
  • 4) construction of atoms

Solution

According to Dalton’s atomic theory chemical reactions involve reorganisation of atoms.

Q24. Which of the following is an element?

  • 1) Sugar
  • 2) Air
  • 3) Oxygen
  • 4) Urea

Solution

Sugar and urea are compounds while air is a mixture. Oxygen is an element having symbol O and exists as diatomic molecule O2.

Q25. What is the molecular mass of a compound if 20% nitrogen is present in it?

Solution

begin mathsize 12px style Assuming space that space straight a space molecule space of space the space compound space contains space at space least space one space nitrogen space atom space then comma space
Molecular space mass space of space the space compound space equals space fraction numerator 100 space cross times space 14 over denominator 20 end fraction space equals space 70
Thus comma space molecular space mass space of space straight a space compound space is space either space 70 space or space multiple space of space it. end style

  Q26. The molecular mass of methane and oxygen are 16 and 32 respectively. If one litre of methane at S.T.P contains N molecules, what will be the number of molecules in 5 L of oxygen at S.T.P?

Solution

According to Avogadro’s law equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules. So 5 L of oxygen at S.T.P contains 5N molecules. 

Q27. Express the following in SI units:

  • A) 93 million miles
  • B) 5 feet 2 inches  

Solution

A) 1 mile = 1.6 Χ 103 m Therefore, 93 million miles = 93 Χ 106 Χ 1.60 Χ 103 m = 1.49 Χ 1011 m

B) 5 feet 2 inches = 5 Χ 12 + 2 = 62 inches = 62 Χ 2.54 Χ 10-2 m = 157.48 Χ 10-2 m = 1.5748 m

Q28. What volume of 10M HCl should be diluted with water to prepare 2.00L of 5M HCl?

Solution

In case of dilution,     M1V1 =  M2V2 10 M HCl 5 M HCl   10 × V1 = 5 × 2.00   

Q29. Two acids H2SO4 and H3PO4 are neutralized separately by the same amount of an alkali when sulphate and dihydrogen orthophosphate are formed respectively. Find the ratio of masses of H2SO4 and H3PO[at mass P = 31] 

Solution

H2SO+ 2NaOH → Na 2SO4 + 2H2O   H3PO4 + NaOH  → NaH2PO4  +  H2O   Equivalent of alkali = 1 g equivalent of H2SO= 1 g equivalent of H3PO4 The two acids must react in the ratio of their equivalent masses.

 begin mathsize 12px style Equivalent space mass space of space straight H subscript 2 SO subscript 4 space equals 98 over 2 space equals space 49
Equivalent space mass space of space straight H subscript 3 PO subscript 4 space equals space 98 over 1 space equals space 98 end style 

Hence ratio of masses of H2SO: H3PO= 49 : 98 =1 : 2.

Q30. What is the mass of 2 litre sample of water containing 25% heavy water (D2O) in it by volume? Density of water is 1.0 g cm-3 where as that of D2O is 1.06 g cm-3.

Solution

begin mathsize 12px style Total space volume space of space straight a space sample space equals space 2 straight L
therefore space Volume space of space straight D subscript 2 straight O space equals space fraction numerator 2000 cross times space 25 over denominator 100 end fraction space equals space 500 space ml
Volume space of space straight H subscript 2 straight O space equals space 2000 space minus space 500 space equals space 1500 space ml
Mass space of space straight H subscript 2 straight O space equals space 1500 space cross times space 1 space equals space 1500 space straight g
Mass space od space straight D subscript 2 straight O space equals space 500 space cross times space 1.06 space equals space 530 space straight g
Total space mass space equals space 1500 space plus space 530
space space space space space space space space space space space space space space space space space space space space space space space space equals space 2030 space straight g space or space 2.03 space kg end style

Q31. How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?

Solution

0.50 mol Na2CO3 means, 0.50 Χ 106 = 53 g of Na2CO3 where as 0.50 M Na2COmeans 1 litre solution of sodium carbonate contains 0.5 moles of sodium carbonate.

Q32. What were the limitations of Dalton’s atomic theory? Explain in detail.

Solution

The main failures of Dalton atomic theory are:

(1) It failed to explain how atoms of different elements differ from each other i.e. It did not tell anything about structure of the atom.

(2) It could not explain how and why atoms different elements combine with each other to form compound atoms or molecules.

(3) It failed to explain the nature of forces that bind together different atoms in a molecule.

(4) It failed to explain Gay Lussac’s law of combining volumes.

(5) It did not make any distinction between ultimate particle of an element that takes part in reactions (atoms) and ultimate particle that has independent existence (molecule).

Q33. 15g of a substance A combines with 20g of a substance B to give 35 g of product C. The law followed in this reaction is:

  • 1) Law of conservation of mass
  • 2) Law of reciprocal proportions
  • 3) Law of definite proportions
  • 4) Law of multiple proportions

Solution

According to the law of conservation of mass, matter can neither be created nor destroyed in a chemical reaction. It always remain conserved.

Q34. Balance the following equations:

  • i) Fe + H2O → Fe3O4 + H2 
  • ii) KMnO+ H2SO4 → K2SO4 + MnSO4 + H2O + O2 
  • iii) I+ HNO3 → HIO+NO2 + H2O
  • iv) Zn + HNO3 → Zn(NO3)2 + N2O + H2O  

Solution

i) 3Fe +4 H2O → Fe3O4 + 4H2 

ii) 4KMnO+ 6 H2SO4 → 2K2SO4 + 4MnSO+ 6H2O + 5O2 

iii) I2 + 10HNO3 → 2HIO3 + 10 NO+ 4H2O

iv) 4Zn + 10 HNO3 → 4Zn (NO3)2 + N2O + 5H2O

Q35. Explain Law of Combining Volumes in detail.

Solution

Gay-Lussac’s Law of Gaseous Volumes (Law of Combining Volumes) This law was given by Gay-Lussac in 1808. He investigated a large number of chemical reactions occurring in gases. As a result of his experiments, Gay-Lussac found that there exists a definite relationship among the volumes of gaseous reactants and products.

Consider the following illustrations. When gases react together to produce gaseous products, the volumes of reactants and products bear a simple whole number ratio with each other, provided volumes are measured at same temperature and pressure.

Example 1: Under identical conditions of temperature and pressure equal volumes, say 1 litre i.e., 1L of each hydrogen and chlorine gases react together to produce double the volume, 2L of hydrogen chloride gas.

Thus, the ratio of volumes is 1 : 1 : 2.

Example 2: Under identical conditions of temperature and pressure 2l of hydrogen gas reacts with 1L of oxygen gas to produce 2L of steam (water vapour)

Thus, the ratio of volumes is 2 : 1 : 2.

Example 3: Under identical conditions of temperature and pressure 1L of nitrogen gas reacts with 3L of hydrogen to produce 2L of ammonia gas.

Here the ratio of volumes is 1 : 3 : 2. Conclusion: All

Examples 4 cited above indicate that during the gaseous reaction, gaseous reactants and products bear a simple ratio of a whole number of volumes with each other.

Q36. How many grams of oxygen (O2) are required to completely react with 0.200 g of hydrogen (H2) to yield water (H2O)?  

Solution

begin mathsize 12px style The space balanced space equation space for space the space reaction space is comma
space space 2 straight H subscript 2 space space plus space space straight O subscript 2 space space rightwards arrow space space 2 straight H subscript 2 straight O
space 2 space mol space space space 1 space mol space space space space space space space 2 space mol
space space space 4 space straight g space space space space space space space space 32 space straight g space space space space space space space space 36 space straight g
4 space straight g space of space straight H subscript 2 space requires space oxygen space equals space 32 space straight g
0.20 space straight g space of space straight H subscript 2 space requires space oxygen space equals space 32 over 4 space cross times space 0.200 space
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space space 1.6 space straight g end style

Q37. How many litres of oxygen are required for the complete combustion of 125 litres of ethylene (C2H2)?

Solution

2C2H2 + 5O2 → 4CO2 + 2H2O 5 Moles of O2 is required for the combustion of 2 moles of C2H2.begin mathsize 11px style Volume space of space straight O subscript 2 space required space equals space 125 space cross times space 5 over 2 space equals space 13.3 end style

Q38. Sapphire weighs 563 carats. If one carat is equal to 200mg, what is the weight of the gemstone in grams?

Solution

Weight of one carat = 200 mg 

begin mathsize 12px style Therefore comma space weight space of space 563 space carats space equals space 200 over 1000 space cross times space 563
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 112.6 space straight g end style

Q39. 2.5 moles of sulphuryl chloride were dissolved in water to produce sulphuric acid and hydrochloric acid. How many moles of KOH will be required to completely neutralize the solution?

Solution

SO2Cl2 + 2H2O → H2SO4 + 2HCl Mol of HCl produced from 2.5 moles of SO2Cl= 5 mol   Mol of H2SOproduced from 2.5 moles of SO2Cl2 = 2.5 mol   H2SO+ 2KOH → K2SO+ 2H2O;   HCl + KOH → KCl + H2O   Now, 5 mol of H2SOrequire KOH = 2.5 mol   Total KOH required = 10 + 2.5 = 12.5 mol. 

Q40. Calculate the molecular formula of a gaseous compound whose 1 volume requires 2 volumes of O2 for combustion and gives 2 volumes of COand 1 volume of Nafter it.

Solution

According to the question,X + 2O→ 2CO2 + N2 Since the product has 2 atoms each of C and N more than that of the reactant, these must be present in the X (reactant). Hence X is C2N2.

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