Showing posts with label Some basic concepts of chemistry. Show all posts
Showing posts with label Some basic concepts of chemistry. Show all posts

IIT JEE PYQ-2000-2017 : Some Basic Concepts of Chemistry

  1. Experimentally it was found that a metal oxide has formula M0.98O.Metal M, is present as M2+and  M3+ in its oxide. Fraction of the metal which exists as M3+  would be: (2013)

    1. 7.01 %

    2. 4.08 %

    3. 6.05 %

    4. 5.08 % 

  2. A gaseous hydrocarbon gives upon combustion 0.72 g . of water and 3.08 g. of CO2.The empirical formula of the hydrocarbon is. (2013)

    1. C2H4

    2. C3H4

    3. C6H5

    4. C7H8


  1. Consider the following reaction,

xMnO4- + yC2O42- + zH+ → xMn2+ +2yCO2+z/2 H2O 

The values of x, y and z in the reaction are, respectively (2007)

    1. 5,2 and 16 

    2. 2,5 and 8 

    3. 2,5 and 16

    4. 5, 2 and 8


  1. In the reaction,

            2AI(s) +6HCl(aq) 2Al3+(aq) + 6Cl- (aq) + 3H2 (g)  (2007)

  1. 11.2 L H2 (g) at STP is produced for every mole HCl (aq) consumed.

  2. 6 L HCl(aq) is consumed for every 3L H2 (g) produced.

  3. 33.6 L H2(g) is produced regardless of temperature and pressure  for every mole Al that reacts.

  4. 67.2 H2(g) at STP is produced for every mole Al that reacts.


  1. The density (in g mL-1) of a 3.60 M sulphuric acid solution that is 29% H2SO4 (molar mass = 98 g mol-1) by mass will be. (2007)

    1. 1.45

    2. 1.64

    3. 1.88

    4. 1.22

  2. Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using diphenylamine as indicator. The number of moles of Mohr’s salt required per mole of dichromate is. (2007)

    1. 3

    2. 4

    3. 5

    4. 6


  1. Density of a 2.05M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is. (2006)

    1. 2.28 mol kg-1

    2. 0.44 mol kg-1

    3. 1.14 mol kg-1

    4. 3.28 mol kg-1


  1. How many moles of magnesium phosphate, Mg3(PO4)2 will contain 0.25 mole of oxygen atoms? (2006)

    1. 1.25 10-2

    2. 2.5 10-2

    3. 0.02

    4. 3.125 10-2


  1. If we consider that ⅙, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of the substance will (2005)

    1. Be a function of the molecular mass of the substance

    2. Remain unchanged

    3. Increase two fold

    4. Decrease twice


  1. Two solutions of a substance (non electrolyte) are mixed in the following manner. 480 ml of 1.5 M first solution + 520 ml of 1.2 M second solution. What Is the Molarity of the final mixture. (2005)

    1. 2.70 M

    2. 1.344 M

    3. 1.50 M

    4. 1.20 M


  1. The pair of the compounds in which both the metals are in the highest possible oxidation state is (2004)

    1. [Fe(CN)6]3- , [Co(CN)6]3-

    2. CrO2Cl2 , MnO4-

    3. TiO3 , MnO2

    4. [Co(CN)6]3- , MnO3


  1. The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide solution for complete neutralization. The organic compound is (2004)

    1. Urea 

    2. Benzamide

    3. Acetamide

    4. Thiourea


  1. To neutralize completely 20 mL of 0.1M aqueous solution of phosphorous acid (H3PO3), the value of 0.1 M aqueous KOH solution required is. (2004)

    1. 40mL

    2. 20mL

    3. 10mL

    4. 60mL


  1. 6.02 1020 molecules of urea are present in 100 mL of its solution. The concentration of urea solution is. (2004)

    1. 0.02M

    2. 0.01M

    3. 0.001M

    4. 0.1M


  1. 25 mL of a solution of barium hydroxide on titration with a 0.1 molar solution of Hydrochloric acid gave a litre value of 35 ml. The molarity of barium hydroxide solution was. (2003)

    1. 0.14

    2. 0.28

    3. 0.35

    4. 0.07


  1. What volume of Hydrogen gas, at 273 K and 1 atm, pressure will be consumed in obtaining 21.6 g of elemental boron (atomic mass = 10.8) from the reduction of boron trichloride by hydrogen ? (2003)

    1. 67.2L

    2. 44.8L

    3. 22.4L

    4. 89.6L


  1. Mixture of X = 0.02 mol of [Co(NH3)5SO4]Br and 0.02 mol of [Co(NH3)5Br]SO4 was prepared in 2 litre of solution.

1 litre of mixture X + excess AgNO3 Y

1 litre of mixture X + excess BaCl2 Z

No. of moles of Y and Z are. (2003)

  1. 0.01,0.01

  2. 0.02,0.01

  3. 0.01,0.02

  4. 0.02,0.02


  1. Number of atoms in 558.5 gram Fe (at. Wt. of Fe = 55.85 g mol-1) is. (2002)

    1. Twice that in 60 g carbon

    2. 6.023 1022

    3. Half that in 8 g He

    4. 558.5 6.023 1023


  1. With the increase of temperature, which of these changes ? (2002)

    1. Molality

    2. Weight fraction of solute

    3. Molarity

    4. Mole fraction


  1. In a compound C,H and N atoms are present in 9:1:3.5 by weight. Molecular weight of the compound is 108. Molecular formula of compound is. (2002)

    1. C2H6N2

    2. C3H4N

    3. C6H8N2

    4. C9H12N3

How to Calculate Gram equivalent mass ? | Formula list | Worksheet

What is Gram Equivalent mass ?

Number of parts by which an element combines with 1 part by mass of hydrogen, or 8 parts by mass of oxygen, or 35.5 parts by mass of chlorine, or one gram equivalent of any other element, is the value of the equivalent mass of the element.

List of formulae:


Practice questions: 

Calculate equivalent mass of Metal in the following compounds : 

  1. Sodium Oxide

  2. Potassium Oxide

  3. Ferrous Oxide 

  4. Magnesium hydride

  5. Aluminium hydride 

Calculate the Equivalent mass of the following Acids and Bases. 

  1. Hydrochloric acid

  2. Magnesium hydroxide

  3. Zinc Oxide 

  4. Cupric hydroxide 

  5. Carbonic acid

Calculate the Equivalent mass of the following salts;

  1. Cadmium chloride 

  2. Zinc Chloride

  3. Mercuric chloride 

  4. Sodium Bromide 

  5. Aluminium Fluoride 


Answers:  

Calculate equivalent mass of Metal in the following compounds : 

  1. Sodium Oxide - 23 g

  2. Potassium Oxide - 39 g

  3. Ferrous Oxide - 27.92 g

  4. Magnesium hydride - 12 g

  5. Aluminium hydride - 9 g 

Calculate the Equivalent mass of the following Acids and Bases. 

  1. Hydrochloric acid - 36.5 g

  2. Magnesium hydroxide - 29.2 g

  3. Zinc hydroxide - 49.75 g

  4. Cupric hydroxide - 48.75 g

  5. Carbonic acid - 31 g

Calculate the Equivalent mass of the following salts;

  1. Cadmium chloride - 55.5 g

  2. Zinc Chloride - 68 g

  3. Mercuric chloride - 135.75 g

  4. Sodium Bromide - 103 g

  5. Aluminium Fluoride - 28 g



Mole Concept | Limiting reagent | Stoichiometry | Practice questions

QUESTION 1:

In this reaction:

CaCO3 + 2HCl ---> CaCl2 + CO2 + H2O

6.088 g CaCO3 reacted with 2.852 g HCl. What mass of CaCO3 remains unreacted?

                                                                                                                                        

SOLUTION :

HCl is Limiting reagent

CaCO3 ⇒ 6.088 g / 100.086 g/mol = 0.0608277 mol

HCl ⇒ 2.852 g / 36.461 g/mol = 0.0782206 mol

Grams of calcium carbonate used:

1 is to 2 as x is to 0.0782206 mol

x = 0.0391103 mol

0.0391103 mol times 100.086 g/mol = 3.914 g

Grams of CaCO3 remaining: 6.088 g minus 3.914 g = 2.174 g

QUESTION 2:Cartoon, Icon, Light Bulb, Symbol

950.0 grams of copper(II) sulfate are reacted with 460.0 grams of zinc metal. (a) What is the theoretical yield of Cu? (b) If 295.8 grams of copper are actually obtained from this reaction, what is the percent yield?

SOLUTION: 

  1. The balanced chemical equation: CuSO4 + Zn ---> ZnSO4 + Cu

According to the balanced chemical equation, calculate the moles of reactants. 

CuSO4 ⇒ 950.0 g / 159.607 g/mol = 5.95212 mol

Zn ⇒ 460.0 g / 65.38 g/mol = 7.03579 mol 

So, CuSO4 is limiting reagent. 

Grams of Cu: 5.95212 mol of Cu is produced (as 1 : 1 molar ratio involved)

5.95212 mol X  63.546 g/mol = 378.2 g

  1. Percent yield : 295.8 g / 378.2 g = 78.21 %


QUESTION 3:

Silicon nitride (Si3N4) is made by combining Si and nitrogen gas (N2) at a high temperature. How much (in g) Si is needed to react with an excess of nitrogen gas to prepare 125 g of silicon nitride if the percent yield of the reaction is 95.0%?

Answer: 79.1 g Si

QUESTION 4:

Use the following reaction: C4H9OH + NaBr + H2SO4 C4H9Br + NaHSO4 + H2O

If 15.0 g of C4H9OH reacts with 22.4 g of NaBr and 32.7 g of H2SO4 to yield 17.1 g of C4H9Br, what is the percent yield of this reaction?

Answer: 61.5 %.

QUESTION 5:

If 4.95 g of ethylene (C2H4) are combusted with 3.25 g of oxygen.

a. What is the limiting reagent?

b. How many grams of CO2 are formed?

Answer: Oxygen and 2.98 g CO2

QUESTION 6:

If you dissolve lead(II) nitrate and potassium iodide in water they will react to form lead(II) iodide and potassium nitrate. 

a.Write the equation for this reaction, then balance the equation. 

b.Calculate the grams of lead(II) iodide that can be produced from 5.00 moles of potassium iodide. 

c.Calculate the grams of lead(II) iodide that can be produced from 75.00 grams of potassium iodide.

Answer:  1.15 x 103 g PbI2 and  104.1 g PbI2 

QUESTION 7:

Write then balance the combustion reaction for propane gas, C3H8. 

a. If 5.00 grams of propane burn completely, what volume of carbon dioxide is produced at STP? 

b. If 75.0 L of steam is produced at STP, what mass of propane must have burned? 

c. If 34.2 grams of propane are completely combusted, how many moles of steam will that produce?

Answer: 7.62 L CO2  ,  36.9 g C3H8 and  3.10 mol steam

QUESTION 8:

If 31.6 grams of magnesium hydroxide is combined with 68.0 mL of 0.725 M HBr, which is the limiting reagent? How many grams of magnesium bromide would be formed? 

Answer: HBr is the limiting reagent and 4.54 g MgBr2

QUESTION 9:

Suppose 316.0 g aluminum sulfide reacts with 493.0 g of water. What mass of the excess reactant remains?

The unbalanced equation is:

Al2S3 + H2O ---> Al(OH)3 + H2S

Answer: Al2S3 is the limiting reagent.227.4685 g of water and 265.5 g excess.

QUESTION 10:

What weight of each substance is present after 0.4500 g of P4O10 and 1.5000 g of PCl5 are reacted completely?

P4O10 + 6PCl5 ---> 10POCl3

Answer: PCl5 is limiting, 0.1092 g of P4O10 remaining,  1.8408 g of  POCl3 produced