XI Chemistry Model Test Questions Chapter # 1 (Test # 1) ….. Stoichiometry

 




XI Chemistry Test Model Questions Chapter # 1 …..Stoichiometry

(Uncertainty in Measurement and Mole Concept)

Test # 1 (Uncertainty in Measurement and Mole Concept)

 

 

Short Question Answers

 

Q1.  Define the following terms:

Exponential notation/scientific notation, standard scientific notation, Significant figures, and Rounding off data


Q2. Define Significant figures. State their rules.


Q3. Define rounding off data. Give valorous rules of rounding off data.


Q4. Write down two rules of finding out numerical value and signs of exponents.


Q5. Define the following terms:

mole, molar volume, molar mass, a.m.u, Avogadro’s number, Stoichiometry, limiting reactant, stoichiometric amount, excess reactant, mass-mass relationship, Avogadro’s law, Gay Lussac’s law, Theoretical Yield/ stoichiometric yield/ calculated yield, Actual Yield/ experimental yield, Percentage Yield

 


Important Mathematical Questions On Uncertainty In Measurement From Book

 

Q1.Express the following numbers in exponential notation

(i) 3652 (3.652 x 103)                 

(ii) 0.0231  (2.31 x 10−2)                    

(iii) 0.000072 (7.2 x 10−5)


Q2. Express the following as the power of ten:

(a) 6782 (6.782 x 103)                

(b) 565.2 (5.652 x102)       

(c) 70,000 (7 x 104)  

(d) .00019 (1.9  x104)


Q3. Express the following in simple numbers

(i) 3.26 x 10−3 ( 0.00326)           

(ii)  1.921 x 102 (192.1)                      

(iii)  1.02 x105 (102000)


Q4. Determine the number of significant figures in each of the following:

(a) 13.01 (Answer; 4 SF) 

(b) 6000 (Answer; 1 SF)       

(c) 0.200 (Answer; 3 SF)            

(d) 0.005 (1 SF)


Q5. Simplify the following using the rule of exponential notation

(i) Add 1.31 x 103 and 3.15 x 102                             (Answer; 1.625 x 103)

(ii) Multiply 7.0  x 1012 and 2.0 x 10−3                (Answer; 1.4 x 1010)

(iii) Divide 6.60  x 108 by 3.20 x 103               (Answer; 2.60 x 105)

(iv) Simplify (3.25  x 104)2                                   (Answer; 1.056 x 109)





IMPORTANT NUMERICALS on Mole Concept from BOOK

 

Q1. Calculate each of the following quantities

(i) Number of moles in 6.4 g of SO2               [Answer; 0.1 mol]

(ii) Mass in gram of 4.5 moles of ethyne (C2H2)    [Answer; 117 g]

(iii) Volume in cm3 of 38.4 g O2 gas at STP      [Answer; 26880 cm3]

(iv) Number of molecules of 126 g water     [Answer; 4.214 x 1024]

(v)  Mass in gram of 4.8 x 1024 atoms of sodium                 [Answer; 183.39 g]

(vi)Number of formula units in 333 g of CaCl2   [18.06 x 1023]

 

Q2. Calculate the number of moles and molecules in

(i) 38 g of carbon disulphide (CS2)             [Answer; 0.5 mole and 3.01 x 1023]               

(ii)68.4 g of sucrose (C12H22O11)                 [Answer; 0.2 mole and 1.24 x 1023]

 

Q3.1.6 g of a sample of gas occupies a volume of 1.12 dm3 at STP. Determine the molar mass of the                 substance. [Answer; 32 g/mol]

 

Q4. What volume of oxygen gas (O2) occupied by 1.5 moles at STP. (Answer; 33.6 dm3)

 

Q5.  Calculate the mass of 3.25 moles of water (H2O).     

(Answer; 58.5 g)

 

Q6. Calculate the number of molecules in 610 g of benzoic acid (C7H6O2). 

(Answer; 3.01 x 1024 molecules)

 

Q7. Calculate the number of moles in 25.5 g of sodium metal.     

(Answer; 1.11 moles)

 

Q8.  Calculate the number of moles in 2.35 x 1025 atoms of aluminium (Al).  

(Answer; 39 moles of Al)

 

Q9. Calculate the mass of 4.39 x 1024 atoms of gold (Au), molar mass of gold is 193 g/mol.

(Answer; 12.365 g of gold)

 

Q10. Graphite is one of the two crystalline forms of carbon which is a constituent component of lead pencils. How many atoms of carbon are there in 360 g of graphite? Also find the number of moles of carbon. 

(Answer; 1.806 x 1025 atoms of C, 30 moles of C)

 



IMPORTANT NUMERICALS on Mole Concept from External Source

 

Q2.  Solve the following:

(i)How many moles are there in 10 g of coke?                                   (Ans; 0.833 mole

(ii)  Calculate the number of moles in 1.505 x 1022 atoms of Mg. 

(Ans; 0.025 mole)

(iii)   Calculate the mass in grams of 3.01 x 1023 atoms of Zinc.      

(Ans; 32.7 g)

(iv)   Calculate the number of atoms in 9.2 grams of Natrium.        

(Ans; 2.408 x 1023 atoms)

(v)  Calculate the number of molecules in 0.8 moles of Glucose.

(Ans;4.816x 1023 molecules)

(vi)   How many molecules are there in 5.4 grams of water?           

(Ans;1.806x 1023 molecules)

(vii)  In a collection of 24 x 1025 molecules of C2H5OH, what is the number of moles? 

(398.6)

 

Q3. The sample of Chlorine gas at S.T.P. has a volume of 800 cm3. Calculate the mass of the sample and the number of molecules in the sample.   

(Ans; 2.2536 g; 2.15 x 1022 molecules)

 

Q4. What is the mass of 3.01 x 1023 molecules of N2? [K.B – 2007]                          

(Ans; 1.4 g)

 

Q5.  Calculate the weight in grams of 3.01 x 1020 molecules of glucose.                  

(Ans; 0.0899 g)


Q6. The atomic mass of sodium is 23 amu. Calculate:       

(a)  The mass of 2.408 x 1023 atoms of Na.

                                                              (Ans; 9.2 g)

(b) The number of moles of Na in 4.6 g.                                                                   

(Ans; 0.2 mole)

 

Q7.  Calculate the mass in grams of 3.01 x 1023 molecules of water.                          

(Ans; 9 g)

 

Q8.Calculate the number of moles, the number of molecules and the volume in cm3 at stp of 0.32 g of:

(i) CH4 gas                                               

(Ans; 0.02 mole, 1.204 X 1022 molecules, 448 cm3)

(ii)SO2 gas                                                

(Ans; 0.005 mole, 3.01X 1021 molecules, 112 cm3)            

 

Q9.Calculate the number of moles and the number of molecules present in 9.0 g of glucose.

(Ans; 0.5 mole, 3.011 X 1023 molecules)

 

Q10. The formula weight of NaOH is 40 amu. Find

(i)Its gram formula weight. (Ans; 40 g)

(ii)Weight of 3 formula unit in amu.  (Ans; 1.99 x 10-22)

(iii)Weight of 60200 formula unit in g. (Ans; 4 x 10-18 g)

(iv)Weight of 6 gram mole.   (Ans; 240 g)

 

Q11.  What is the mass of following in gram

(i)  One atom of Silver  (Ans: 1.79 x 10–22 g

(ii) One atom of Cu  (Ans: 1.05 x 10–22 g)

(ii)  One atom of Zn (Ans: 5.43 x 10–22 g)

(iv)One atom of Gold  (Ans: 3.27 x 10–22 g

(v)  One atom of Iron  (Ans: 9.3 x 10–22 g

(v) One atom of Carbon (Ans: 9.3 x 10–22 g)

 

Q12. 10 g of H2SO4 has been dissolved in excess of water to dissociate it completely into its ions. Calculate

(i)Number of molecules in 10 g of H2SO

(Ans: 6.14 x 1022 molecules)

(ii)Number of positive ions                                           

(Ans: 1.2 x 1023 molecules)

[Hint; No of positive  ions = Number of molecules x basicity of Acid]

 





 

MCQs on Chapter # 1 Section I …. Uncertainty in Measurement

 









Answers Key MCQs on Chapter # 1 Section I …. Uncertainty in Measurement



 

Text Book MCQs

 

Q1. Choose the best right answer from the given options 

1) If the volume occupied by oxygen gas (O2) at STP is 44.8dm3, the number of molecules of O2 in the vessels are:

(a) 3.01 x 1023                                  

(b) 6.02 x 1023                                  

(c) 12.04 x 1023               

(d) 24.08 x 1023

 

2)  The number of carbon atoms in 1 mole of sugar (C12H22O11) are approximately:

(a) 6 x 1023                                        

(b) 24 x 1023                                     

(c) 60x1023                        

(d) 72x1023

 

3) In the reaction 2Na + 2H2 O  → 2NaOH + H2, if 23g of Na reacts with excess of water ,the volume of hydrogen gas (H2) liberated at STP should be:

(a) 11.2 dm3                                      

(b) 22.4 dm3                                      

(c) 33.6 dm3                      

(d) 44.8 dm3

 

4)Which of the following sample of substances contains the same number of atoms as that of 20g   calcium?

(a) 16g S                                             

(b) 20g C                                            

(c) 19g K                             

(d) 24g Mg

 

5) Number of atoms in 60g carbon are:

(a) 3.01 x 1023                                  

(b) 3.01 x 1023                                  

(c) 6.02 x 1023                  

(d) 6.02 x 1024

 

6)Maximum number of molecules present in the following sample of gas:

(a) 100g O2                                        

(b) 100g CH4                                     

(c) 100g CO2                     

(d) 100g Cl2

 

7) Which of the following statement is incorrect?

(a) The mass of 1 mole Cl2 gas is 35.5g

(b)One mole of H2 gas contains 6.02x1023 molecules of H2

(c) Number of atoms in 23g Na and 24g Mg are equal

(d) One moles of O2 at S.T.P occupy 22.4dm3 volume

 

8) For Avogadro’s number, this statement is incorrect:

(a)  It is the no. of particles in one mole of any substances

(b) Its numerical value is 6.02 x 1023

(c)Its value change if temperature increases

(d)Its value change if number of moles increases

 

9)  Generally actual yield is:

(a)  Greater than theoretical yield

(b) Less than theoretical yield

(c) Equal to the theoretical yield

(d) Sometimes greater and sometimes less than theoretical yield

 

10) The minimum number of moles are present in:

(a) 1dm3 of methane gas at STP    

 (b) 5dm3 of helium gas at STP

(c) 10dm3 of hydrogen gas at STP                         

(d) 22.4dm3 of chlorine gas at STP

 

 






Answer 

Significant Figures/Significant Digits

The Statistically Significant digits or Meaningful or Reliable digits of a number known with certainty in a measured (calculated) quantity which are needed to express the precision of the measurement are known as significant figures. The certain digits of a measured quantity plus one uncertain rightmost last digit are counted as significant figures.




Exponential Notation/ Scientific Notation

The short hand expression of a very large or a very small number by means of exponents is called exponential notation or scientific notation. Scientific notation is a product of co-efficient number (digit term) and 10 raised to some power (exponential term).




Standard Scientific Notation

Standard Scientific Notation is one in which decimal point is after one digit of co-efficient number. For Example

4.56 x 106 is a standard scientific notation while 45.6 x 105 is not a standard scientific notation”.

 

Rounding Off Data

To round off means to reduce a number to the desired significant figures. The procedure of dropping non-significant digits of a data to reduce a number to the required significant digit and adjusting the last digit reported is called Rounding Off Data.  









Answer

Definition                                                                                     

The Statistically Significant digits or Meaningful or Reliable digits of a number known with certainty in a measured (calculated) quantity which are needed to express the precision of the measurement are known as significant figures. The certain digits of a measured quantity plus one uncertain rightmost last digit are counted as significant figures.






Guideline Rules for Determining the Significant Figures                        

 

1.  Non-zero digits (any digit that is not zero i.e. 1-9 integers) are all significant.

                e.g.

(i)67 m has two significant  figures.                             

(ii)  435 kg has three significant figures.

 

2.   Captive Zeros (i.e. zeros placed between two non-zero digits) always count as significant.

                

e.g.

 (i)  4090 mg has three significant figures. 

(ii) 28.073 g has five significant figures.

 

3.  Leading Zeros (i.e. zeros precede the nonzero digits) are not significant.

e.g.

(i)  0.0006 g has 1 significant figures.           

(ii) 0.00027 g has 2 significant figures.

 

4.  Trailing zeros (i.e. zeros at the right end of the number) right of decimal point are always significant

 (i) 3.00m has three significant figures.       

(ii) 5.900m has four significant figures.

 

5. Trailing zeros in numbers with no decimal point are not significant Zeros

(i) 140 miles has 2 significant figures.                       

(ii) 786000 calories has 3 significant figures.




Q3. Define rounding off data. Give valorous rules of rounding off data. 

Answer 

Definition of Rounding Off Data

To round off means to reduce a number to the desired significant figures. The procedure of dropping non-significant digits of a data to reduce a number to the required significant digit and adjusting the last digit reported is called Rounding Off Data. 

 

Rules for Rounding Off Data

1.    If last dropping digit is greater than five, then the last remaining digit to be retained is increased by one unit. This is called ROUNDING UP.

e.g.

5.768  is rounded off to 5.77 to 3 significant digits.

5.768  is rounded off to 5.8   to 2 significant digits.

 

2.   If last dropping digit is less than five, then the last remaining digit will remain unchanged. This is called ROUNDING DOWN.     

e.g.

5.734  is rounded off to 5.73 to 3 significant digits.

5.734  is rounded off to 5.7   to 2 significant digits.

73 rounded down to the nearest ten is 70, because 73 is closer to 70 than to 80.

 

3.  If last dropping digit is exactly five, then the last remaining digit is increased by one unit if it is odd and remain unchanged if it is even e.g.

7.865 is rounded off to 7.86 to 3 significant digits          [L.R.D. = even]

8.775 is rounded off to 8.78 to 3 significant digits          [L.R.D. = odd]






Answer

Rule I for determination of Numerical Value of Exponents


Exponent is positive, when decimal point is shifted towards left. The exponent is numerically equal to the number of places the decimal point has been moved. The numbers which are greater than one have positive exponents. Such notations may be represented by a general formula 10n where n is the number of zeros. e.g.





Rule II for determination of Numerical Value of Exponents

Exponent is negative, when decimal point is shifted to right. The exponent is equal numerically to the number of places the decimal point has been moved. The numbers which are less than one have negative exponents. Such notations may be represented by a general formula 10n+1 where n is the number of zeros. e.g.





Q5.Simplify the following using the rule of exponential notation

(i)       Add 1.31 x 103 and 3.15 x 10

(ii)     Multiply 7.0  x 1012 and 2.0 x 10−3

(iii)    Divide 6.60  x 108 by 3.20 x 103 

(iv)    Simplify (3.25  x 104)2

(v)       

Answer

 

(i) Add 1.31 x 103 and 3.15 x 10

Before addition, the value 3.15 x10is converted into 0.315 x 103 by placing decimal point to the left to get same exponents of 10 as that of 1.31 x 103. Then coefficients of both values are added to get the result.

   1.31 x 103

+ 0.315 x 103

                                   1.625 x 103

 

(ii)  Multiply 7.0  x 1012 and 2.0 x 10−3

= (7.0)  (2.0) x 1012−3

= 14 x 109

= 1.4 x 1010

 

(iii)       Divide 6.60  x 108 by 3.20 x 103 

6.60  x 108/3.20 x 103  = 2.60 x 108−3 2.60 x 105

 

(iv)   Simplify (3.25  x 104)2

Here digit term is 3.25 and exponent is 104 and both are multiplied by whole power of the figure i.e. 2 to get the answer

 

(3.25 )2 x 104x2 = 10.56 x 101.056 x 109

 















a.m.u

An atomic mass unit (abbreviated as a.m.u.) is a physiochemical constant and it is defined as one twelfth (1/12) of the mass of a single atom (the most abundant lightest isotope) of carbon-12 (12C). It is the mass of 1/12 part of carbon-12 atom. It is equal to one over Avogadro’s number when expressed in gram.

 1/2th weight of carbon–12=  1 a.m.u. =   Mass of one H atom

1 a.m.u (atomic mass unit)=1.66 x 10–24 g=   1.66 x 10–27 kg

 

molar mass

The relative atomic mass or molecular mass or formula mass of a substance expressed in gram is called its molar mass. In other words, the mass in gram of 1 mole of units (such as atoms, molecules or ions) of a substance is called its molar mass. The unit of molar mass is gram or gram/mol.

 

Mole

Mole symbolized as mol is the S.I. base unit of chemical amount of substance. A mole of a substance can be defined as the atomic mass or molecular mass or formula mass of a substance (element or compound) expressed in gram containing 6.02 x 1023 particles (Atoms, molecules or ions).

Avogadro’s number

“The number of elementary particles (atoms, molecules, ions or formula units) contained in one mole (i.e. gram atomic or gram molecular or gram formula mass) of a substance (element or compound) is found to be 6.02 x 1023  called Avogadro’s number or Avogadro’s constant.  It is denoted as NA.” It is the number of particles per mole.

6.02 x 1023particles=1 mole of any substance


molar volume

It is the volume occupied by one mole of any gas at s.t.p or r.t.p. 22.4 dm3 (Litre) or 0.0224 m3 or 22400 cm3 (ml) of any gas at STP is referred to as Molar Volume. The unit of molar volume is dm3 mol-1 (or cm3 mol-1 or m3 mol-1).

Stoichiometry  

The branch of chemistry which deals with the study of quantitative relationships between the amounts of reactants and products of a chemical reaction as given by a balanced chemical equation is called Stoichiometry.  It is a very mathematical part of chemistry.

 

Stoichiometric Amounts

The amount of reactants and products in balanced chemical equations are called Stoichiometric Amounts. According to Law of Conservation of Mass, the weight of reactants is equal to weight of products.

 

Mass–Mass Relationship/Calculation based on Mass–Mass Relationship

Such relationships are helpful in determining unknown mass of a reactant or product from the given mass of reactant or product with the help of balanced chemical equation.

 

Mass–Volume Relationship/Calculation based on Mass–Volume Relationship

The Mass–Volume relationship or volume-mass calculations are used when either one of the reactant or product is in gaseous state. The mass – volume relationships are useful is determining the unknown volume or mass of reactants or products from a known volume or mass of some substances in a chemical reaction. It is based on Avogadro’s law which states that 1 mole (1 g/mole) of every gas always occupies 22.4 dm3 (22.4 litre) or 22400 cm3 or 0.0224 m3 at S.T.P. 

 

Volume – Volume Relationship

Volume-Volume Relationship is used for determination of volumes of gases. The volume-volume relationships are useful in determining unknown volume of reactants or products from a known volume of reactants or products.  It is based on Gay-Lussac’s Law of Combining Volumes

 

Gay-Lussac’s Law of Combining Volumes

Gases combine or form in chemical reactions in the ratio of simple whole numbers by volumes provided at same temperature and pressure”. The ratio of the volumes of gases is same as the ratios of their molecules in a balanced equation

 

Limiting Reactants

The reactant which is entirely consumed first during chemical reaction is called limiting reactant or limiting reagent. Mathematically, limiting reactant is that which gives least number of moles of product

 

Excess Reactant

The reactant that is not completely consumed is called as excess reactant.

 

 

Theoretical Yield/ stoichiometric yield/ calculated yield

The maximum amount of the product calculated from the balanced chemical equation by using its limiting reactant (given amount of reactant) is known as theoretical yield or stoichiometric yield or calculated yield or expected yield.

 

Actual Yield/ experimental yield

The actual amount of product which is formed in an experiment is called practical or actual yield. The amount of the products actually obtained from a given amount of the reactant in a chemical reaction experimentally is called the actual yield or experimental yield of that reaction.

 

Percentage Yield

The efficiency of a chemical reaction can be checked by calculating its percentage yield which is expressed by comparing the actual yield and theoretical yields. The ratio of practical yield to theoretical yield is referred as percent yield








































































































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