1. Solution :
Mr of H2C2O4 = (6 x Ar H) + (2 x Ar C) + (6 x Ar O)
= (6 x 1) + (2 x 12) + (6 x 16)
= 126 amu
Molar mass of H2C2O4 = 126 g/mol
Mole of H2C2O4= Massa of H2C2O4
Molar mass H2C2O4
= 1,26 gram
126 /mol
= 0,01
Jawab B (Ebtanas 1993)
2. Solution :
Mr of Al2(SO4)3= (2 x Ar Al) + (3 x Ar S) + (12 x Ar O)
= (2 x 27) + (3 x 32) + (12 x 16)
= 342 amu
Molar mass of Al2(SO4)3 = 342 g/mol
Mass of Al2(SO4)3 = Mole of Al2(SO4)3 x Molar mass of Al2(SO4)3
= 0,2 mol x 342 g/mol
= 68,4 g/mol
Jawab D (Ebtanas 1994)
3. Solution :
Mr of O2 = 2 x Ar O
= 2 x 16
= 32 amu
Molar mass of O2 = 32 g/mol
Mole of O2 = Mass of O2
Molar mass of O2
= 8 g
32 g/mol
= 0,25 mol
Molecules of O2 = 0,25 mol x 6,022 . 1023 molekul/mol
= 1,505 . 1023 molekul
Jawab A (Ebtanas 1994)
4. Solution :
Volume of N2 = Mole of N2 x Volume of STP
= 0,5 mol x 22,4 L/mol
= 11,2 L
Jawaban B (Ebtanas 2000)
5. Solution :
Atoms of H2 = 2 x the amount of molecule H2
= 2 x mol H2 x Avogadro’s number
= 2 x0,5 mol x 6,022 . 1023 atoms/mol
= 6,02 x 1023 atoms
Jawaban D (Ebtanas 2000)
6. Solution :
Mole of gas = V gas
V STP
= 2,8 L
22,4 L/mol
Mole of gas also can be calculated by :
Mole of gas = Mass of gas
Molar mass of gas
= 4,25 g/ Molar mass of gas
So, we can arrange those value above to be
2,8 L = 4,25 g
22,4 L/mol Molar mass of gas
Molar mass of gas = 4,25 g x 22,4 L/mol
2,8 L
= 34 /mol
Mr of gas is 34 amu
Jawaban E (UMPTN 1993)
7. Solution :
Mole of CO = 2,8 g
28 g/mol
= 0,1 mol
Mole of O2 = 3,2 g
32 g/mol
= 0,1 mol
Mole of O2 = Mole of CO
Find mole of another gas with the same way as the calculation above.
Jawaban : A
8. Solution :
Mole of CH4 = 2,8 L
22,4 L/mol
= 0,125 mol
Atoms of CH4 = 5 x the amount of molecule CH4
= 5 x Mole of CH4 x 6,022 . 1023 molecule/mol
= 5 x 0,125 x 6,022 . 1023
= 3,76375 . 1023
Find the amount of atoms of another compound with the same way as the calculation above. Then you will know that C2H2 has greater number than the other.
Jawaban : E (UMPTN 1998)
9. Solution :
Atoms O in H2SO4 = 4 x the number of molecules H2SO4
= 4 x mole H2SO4 x avogadro’s number
= 4 x 2 mol x 6,022 . 1023atoms/mol
= 48,176 . 1023 atoms
Atom O in KMnO4= 4 x the number of molecules KMnO4
= 4 x mole KMnO4x avogadro’s number
= 4 x 2 mol x 6,022 . 1023 atoms/mol
= 48,176 . 1023 atoms
Jawaban : C
10. Solution :
The amount of molecules = mol x Avogadro’s number
= mol x L = 10 L
Jawaban : B (SPMB 2005)
11. Solution :
Ar X = mass of 1 atom X
1/12 x massa atom C
= 2,67 . 10-23 g
1/12 x 2 . 10-23
= 16 amu
Jawaban A (EBTANAS 1998)
12. Solution :
Mass of one molecule H2O = 18 g/mol
6,022 x 1023 molecules/mol
= 3 x 10-23
Mass of one molecule C-12 = 12 g/mol
6,022 x 1023 molecules/mol
= 2 x 10-23
Mr of H2O = Mass of one molecule H2O
1/100 x Mass of one molecule C-12
= 3 x 10-23
1/100 x 2 x 10-23
= 150 amu
Jawaban B (Proyek Perintis 1982)
13. Solution :
Mole of X = The amount of atoms X
Avogadro’s number
= 2,4 x 1023 atoms
6,0 x 1023 atoms/mol
= 0,4 mol
Molar mass of X = Massa of X
Mole of X
= 60 g
0,4 mol
= 150 g/mol
Ar of X = 150 amu
Jawaban : C (Sipenmaru 1988)
14. Solution :
Mol N2 = 4 g
28 g/mol
= 0,143 mol
Mol H2 = 2 g
2 g/mol
= 1 mol
Mol CH4 = 4 g
16 g/mol
= 0,25 mol
Mol O2 = 8 g
32 g/mol
= 0,25 mol
Mol CO2 = 16 g
44 g/mol
= 0,36 mol
The compound which has the smallest number of mole, has the smallest number of molecule. So N2 has the smallest number of atoms.
Jawaban : D
15. Solution :
Mr of H2S = (2 x Ar H) + (Ar S)
= 2 + 32 = 34 amu
Molar mass of H2S = 34 g/mol
Mass of H2S = Mole of H2S x Molar mass of H2S
= 0,4 mol x 34 g/mol
= 13,6 g
Mole of S = mole of H2S = 0,4 mol
Mass of S = 0,4 mol x 32 g/mol = 12,8 g
Jawaban : C
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