History of the Mole


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History of the Mole

  Amadeo Avogadro


The number of objects in one mole, that is, 6.02 x 1023, is commonly referred to as Avogadro's number. Amadeo Avogadro was an Italian physics professor who proposed in 1811 that equal volumes of different gases at the same temperature contain equal numbers of molecules. About fifty years later, an Italian scientist named Stanislao Cannizzaro used Avogadro's hypothesis to develop a set of atomic weights for the known elements by comparing the masses of equal volumes of gas. Building on this work, an Austrian high school teacher named Johann Josef Loschmidt calculated the size of a molecule of air in 1865, and thus developed an estimate for the number of molecules in a given volume of air. While these early estimates have since been refined, they led to the concept of the mole - that is, the theory that in a defined mass of an element (its atomic weight) there is a precise number of atoms: Avogadro's number.
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Answer Key (Exercise Problem 2)


Siska Sintia Depi

1.        Solution :
Mole of NO = Mass of NO
                       Molar mass of NO
                    = 7,5 g        
                       30 g/mol
                    = 0,25 mol
n1/V­1 = n2/V2
0,25 mol     =   n of X
10 L                  1 L
Mole of X = 0,25 mol x 1 L  = 0,025 mol
                    10 L
Molar mass of X = Mass of X
                              Mole of X
                           = 1,5 g/0,025 mol = 60 g/mol
Mr of X = 60 amu
Jawaban B(SPMB UNPAD 2005)
2.       Solution :
Take the greatest value of Mr of Nitrogen
                                              Mr substance
In (NH4)2SO4 ,  28/142 = 0,197
In (NH4)3PO4, 42/150 = 0,28
In (NH2)2CO, 28/60 = 0,467
In NH4NO3, 14/80 = 0,175
In NaNO3, 14/85 = 0,165
Jawaban C (SPMB 2004)
3.       Solution :
Mole of NO = 5 gram
                       30 g/mol
                    = 0,167 mol
Mole of NO2 = 6 gram
                       46 g/mol
                    = 0,14 mol
Mole of N2O3 = 7 gram
                         76 g/mol
                      = 0,09 mol
Mole of N3O4 = 8 gram
                          92 g/mol
                       = 0,087 mol
Mole of N2O5 = 10 gram
                         108 g/mol
                       = 0,093 mol
Jawaban A (SPMB 2004)
4.       Solution :
Mole = 2,4 x 1022 molecule
            6,022 x 1023 molecule/mol
        = 0,04 mol
Molar mass = 1,6 g / 0,04 mol =40 g/mol
Mr = 40 amu        
Jawaban E (USM Kedokteran UNJANI 2004)
5.       Solution :
Mr of C4H10 = (10 x Ar H) + (4 x Ar C)
                    = (10) + (4 x 12)
                    = 58 amu
Mr of CO2    = (2 x Ar O) + (Ar C)
                    = (2 x 16) + (12)
                    = 44 amu
Mr of NO2   = (Ar N) + (2 x Ar O)
                    = (14) + (2 x 16)
                    = 46 amu
Mr of C3H4  = (4 x Ar H) + (3 x Ar C)
                    = (4 x 1) + (3 x 12)
                    = 40 amu
Mr of SO2    = (Ar S) + (2 x Ar O)
                    = (32) + (2 x 16)
                    = 64 amu
Because mass of each compound is same, the smallest Mr has the biggest number of mole. The biggest number of mole has the biggest volume. 
Jawaban D (USM Kedokteran UNJANI 2004)
6.       Solution :
Molar mass NO2 = 46 g/mol
Mole of NO2 = 4,6 g         = 0,1 mol
                        46 g/mol 
0,1 mol x 6,022 x 1023 = 6,022 x 1022
Jawaban C
7.       Solution :
In 1 mol Fe2(SO4)3 contain 3 mol of Sulfur, so in 2 mol Fe2(SO4)3 contain 6 mol of Sulfur
Jawaban E
8.       Soluton :
The total number of atoms is 3
The amount of atoms = 3 x 1,2 x 1020 molecule of CO2
                                     = 3,6 x 1020
Jawaban  C
9.       Solution :
Mr of N2­O = (2 x Ar N)+ (Ar O)
              = (2 x 14) + (16)
              = 44 amu
Molar mass of N2­O = 44 g/mol
Mole of N2­O = 3,01 .1022 molecule
                        6,02 .1023 molecule/mol
                     = 0,05 mol
Mass of N2­O = mole x molar mass
                      = 0,05 mol x 44 g/mol
                      = 2,2 gram
Jawaban B
10.   Solution :
1 mol = 6.022 x 1023 atoms
Mole of Na = 23 g        = 1 mol, Na has 6.022 x 1023 atoms
                      23 g/mol
Jawaban A
11.      Solution :
Mole of O2 = V O2
                     V STP
                  = 1,12 L
                     22,4 L/ mol
                  = 0,05 mol
Mass of O­2 = Mole of O2 x Molar mass of O2
                   = 0,05 mol x 32 g/mol
                   = 1,6 gram
Jawaban C
12.    Solution :
Mole of X   = V of X
                     V STP
                  = 5,6 L
                     22,4 L/ mol
                  = 0,25 mol
Molar mass of X = Mass of X
                              Mole of X
                           = 10 g         
                              0,25 mol
                           = 40 g/mol
Jawaban C
13.    Solution :
n1/V­1 = n2/V2
n1 = n2 , at the same temperature
Mole of B gas = Mole of A gas
Mole of B gas = 6 x 1022 molecule
                         6 x 1023 molecule/mol
                                         = 0,1 mol
Mole of A gas = 0,1 mol
Molar mass of A = 3 g / 0,1 mol = 30 g/mol                   
Jawaban D
14.    Solution :
Mole of CH­4 = 8 g          = 0,5 mol
                        16 g/mol
Mole of CH­4 = 1,6 g          = 0,05 mol
                        32 g/mol
n1/V­1 = n2/V2
0,5 mol     =      0,05 mol
    x L                  1 L
x L = 0,5 mol x 1 L  =  10 L
         0,05 mol
Jawaban B
15.    Solution :
Molar mass of NO = 30 g/mol
Mole of NO = 10 g       
                       30g/mol
PV = nRT
V = nRT/P
   = 1/3 mol x 0,082L atm mol-1 K-1 x (27 + 273)K    = 8,2 L
                   1 atm
Jawaban A


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Exercise Problem 2


Siska Sintia Depi

1.        At certain temperature and pressure, 1 liter of X gas has mass 1,5 gram. If in that condition the mass of 10 L of NO gas (Mr =30) is 7,5 gram, so relative molecular mass of X gas is…
A.     67,2
B.      60
C.      22,6
D.     30
E.      26,5
2.       In the following fertilizer, which one is contain the greater number of Nitrogen?
A.     (NH4)2SO4 (Mr = 142)
B.      (NH4)3PO4 (Mr = 150)
C.      (NH2)2CO (Mr = 60)
D.     NH4NO3 (Mr = 80)
E.      NaNO3 (Mr = 85)
3.       In the following substance which one is contain the greater amount of molecule…
A.     5 gram of NO(Mr = 30)
B.      6 gram NO2(Mr = 46)
C.      7 gram N2O3(Mr = 76)
D.     8 gram N3O4(Mr = 92)
E.      10 gram N2O5 (Mr = 108)
4.       1,6 gram of any substance contain 2,4 x 1022 molecule. Mr of this substance is…
A.     16
B.      24
C.      32
D.     36
E.      40
5.       At certain temperature and pressure, which one of the following substance has the greater volume in 1 gram of substance…
A.     C4H10
B.      CO2
C.      NO2
D.     C3H4
E.      SO2
6.       In 4,6 gram NO2 (Ar N = 14; O = 16) is contained ... molecule NO2.
A.      6,02 x 1023
B.      3,01 x 1023
C.      6,02 x 1022
D.      3,01 x 1020
E.       1,20 x 1020
7.       The number of mol of sulfur in 2 mol Fe2(SO4)3 is....
A.     1 mol
B.      2 mol
C.      3 mol
D.     5 mol
E.      6 mol
8.       An closed cylinder contain 1,2 x 1020 molecule of CO2  gas at certain T and P.  The amount of atoms which contain in this cylinder is....
A.     4 x 1019
B.      8 x 1019
C.      3,6 x 1020
D.     3,6 x 1021
E.      7,2 x 1021
9.       X gram of gas N2O (Ar N = 14, O = 16) is put into closed vessel. If the amount of particle of gas is 3,01 x 1022 molecule, so the mass of gas is....
A.     1,1 gram
B.      2,2 gram
C.      3,3 gram
D.     4,4 gram
E.      5,5 gram
10.   In the following unsure, unsure which contain 6.02 x 1023 atoms is....
A.     23 g Na, Ar = 23
B.      28 g Mg, Ar = 24
C.      24 g Ca, Ar = 40
D.     42 g Kr, Ar = 40
E.      78 g K, Ar = 39
11.      At 0oC and 1 atm 1,12 L O2 gas has mass... gram. (Ar O = 16)
A.     0,8
B.      1,2
C.      1,6
D.     2,1
E.      2,5
12.    A 5,6 liter of X gas at STP has mass 10 gram. The molar mass of X gas is....
A.     20 gram/mol
B.      30 gram/mol
C.      40 gram/mol
D.     50 gram/mol
E.      60 gram/mol
13.    There are 2 vessel which have the same volume. The first vessel contain 3 gram of A gas, and the second vessel contain 6 x 1022 molecule B gas. If 2 vessel has same temperature and pressure, so molar mass of A gas is...(1 mol = 6 x 1023 molecule)
A.     90 g/mol
B.      75 g/mol
C.      60 g/mol
D.     30 g/mol
E.      25 g/mol
14.    Methane gas, CH4 which has mass 8 gram (Mr = 16) and volume = x L is measured at same T and P as O2 gas which each liter of the O2 gas has  mass = 1,6 gram. Volume of CH4 gas is....
A.     5 L
B.      10 L
C.      15 L
D.     20L
E.      25 L
15.    At 27oC and 1 atm NO gas (Ar N = 14, O = 16) which has mass 10 gram, has volume…
A.     8,2 L
B.      12,0 L
C.      16,4 L
D.     24,5 L
E.      30,7 L
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