konsep mol
DESCRIPTION
KIMIA SMK ASYIKTRANSCRIPT
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KONSEP MOLTri Goesema Putra S, M.Pd
By KIMIA SMK ASYIK
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Jembatan Mol
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Mol1. Satuan banyaknya zat 2. Jumlah zat yang terdapat dalam
6,02 x 1023 partikel3. Volume dari 1 mol suatu zat dalam
keadaan gas. 4. Kosentrasi Molaritas dalam 1 liter
larutan
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Mol1. Satuan banyaknya zat.
Rumus: n = m / Mr
m = n x MrDimana: n : molm : massa (gram)Mr : Massa Molekul RelatifAr : Massa Atom Relatif
m
n Mr
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Mol2. Jumlah zat yang terdapat dalam
6,02 x 1023 partikel.Rumus:
n = JP / 6,02 x 1023
JP = n x 6,02 x 1023
Dimana: n : Mol JP : Jumlah partikel Bilangan Avogadro (L) : 6,02 x 1023
JP
n L
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Mol3. Volume dari 1 mol suatu zat dalam
keadaan gas.Rumus:
n = V / 22,4 V = n x 22,4
Dimana: n : mol V : Volume (liter)
V
n 22,4
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Mol4. Hubungan mol dengan kosentrasi larutan.
Rumus: M = n / V n = M x VDimana: n : Mol V : Volume (liter)M : Molaritas
n
M V(liter)
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Contoh 1
Hitung massa dari 0.25 mol dari butana (C4H10).
Latihan
1. Hitung massa dari 2.5 mol Kalsium Karbonat (CaCO3).
2. Hitung massa dari 0,1 mol dari Amonium sulfat (NH4)
2SO4,
Mr = 100 g
massa = 2.5 x 100 = 250 g
Mr = 132 g
massa = 0.1 x 132 = 13.2g
Step 3:- Gunakan rumus “SEGITIGA AJAIB”
massa = mol x Mr
massa = 0.25 x 58
massa = 14.5 g
Step 2:- Hitung Mr senyawa butana: (4 x 12) + (10 x 1) = 58 g
Step 1 :- Tulis rumus dari butana C4H10
m
n Mr
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m
n Mr
HITUNGAN MOL
𝑛=𝑚𝑀𝑟
massa =
HAPUS
mol = Mr =
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m
n Ar
HITUNGAN MOL
𝑛=𝑚𝐴𝑟
massa =
HAPUS
mol = Ar =
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JP
n L
HITUNGAN MOL
𝑛=𝐽𝑃
6,02𝑥1023
6,02 x 1023
Tuliskan angka matematika L seperti:6.02 x 1023 adalah 6.02E+23
JP =
HAPUS
mol =
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JP
n 22.4
HITUNGAN MOL
𝑛=𝑉22,4
22.4
V =
HAPUS
mol =
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n
M V
HITUNGAN MOL
𝑀=𝑛𝑉
mol =
HAPUS
M = V =
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LATIHAN KONSEP MOL ...
KOREKSI
HAPUS
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