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Pertanyaan

hitung normalitas berikut
a. 4,5 gram alumunium klorida dalam 150 ml larutan
b. 15 gram barium hidroksida dalan 1,15 L larutan
c. 12 gram kalium klorida dalam 200 ml larutan
d. 400 mg calsium sulfat dalam 500 ml larutan
e. 0,5 gram asam phospat dalam 300 ml larutan
f. 125 mg kalium iodida dalam 2,5 L larutan

2. Hitunglah berapa gram natrium karbonat murni diperlukukan untuk membuat 250 ml larutan 0,150 N
3. Hitung volume yang dibutuhkan untuk melarutkan 45 mg natrium hidroksida untuk membuat larutan dengan konsentrasi 0,1 N

1 Jawaban

  • [a]
    AlCl₃
    Mr = 133,5
    ek = 3
    wt = 4,5 gram
    V = 150 mL
    N = ...................?

    BE = Mr/ek
    = 133,5/3
    = 44,5

    N = wt/BE x 1000/V
    = 4,5/44,5 x 1000/150
    = 0,1 x 6,67
    = 0,667 normal

    [b]
    Ba(OH)₂
    Mr = 171
    ek = 2
    wt = 15 gram
    V = 1,15 L =1150 mL
    N = ........................?

    BE = Mr/ek
    = 171/2
    = 85,5

    N = wt/BE x 1000/V
    = 15/85,5 x 1000/1150
    = 0,175 x 0,87
    = 0,152 normal

    [c]
    KCl
    Mr = 75
    ek = 1
    wt = 12 gram
    V = 200 mL
    N = ........................?

    BE = Mr/ek
    = 75/1
    = 75

    N = wt/BE x 1000/V
    = 12/75 x 1000/200
    = 0,8 normal

    [d]
    CaSO₄
    Mr = 136
    ek = 1
    wt = 400 mg = 0,4 gram
    V = 500 mL
    N = ........................?

    BE = Mr/ek
    = 136/1
    = 136

    N = wt/BE x 1000/V
    = 0,4/136 x 1000/500
    = 0,02 normal

    [e]
    H₃PO₄
    Mr = 98
    ek = 3
    wt = 0,5 gram
    V = 300 mL
    N = ........................?

    BE = Mr/ek
    = 98/3
    = 32,67

    N = wt/BE x 1000/V
    = 0,5/136 x 1000/300
    = 0,012 normal

    [f]
    KI
    Mr = 166
    ek = 1
    wt = 125 mg = 0,125 gram
    V = 2,5 L = 2500 mL
    N = ........................?

    BE = Mr/ek
    = 166/1
    = 166

    N = wt/BE x 1000/V
    = 0,125/166 x 1000/2500
    = 0,003 normal

    [2]
    NaCO₃
    ek = 2
    Mr = 106
    V = 250 mL
    N = 0,15 normal
    wt = ...........................?

    BE = Mr/ek
    = 106/2
    = 53

    N = wt/BE x 1000/V
    0,15 = wt/53 x 1000/250
    0,15 x 53 = wt x 4
    wt = 7,95/4
    wt = 1,99 gram

    [3]
    NaOH
    ek = 1
    Mr = 40
    wt = 45 mg = 0,045 gram
    N = 0,15 normal
    V = ...........................?

    BE = Mr/ek
    = 40/1
    = 40

    N = wt/BE x 1000/V
    0,1 = 0,045/40 x 1000/V
    0,1 = 1,125/V
    V = 1,125/0,1
    V = 11,25 mL

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