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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