Pola Soal Fraksi Mol Dan Pembahasannya

Nah topik kita kali ini masih membahas tentang bagaimana menyatakan konsentrasi suatu larutan, dimana salah satunya dengan fraksi mol. Apa itu Fraksi Mol ? Fraksi mol yaitu suatu besaran fokus larutan yang menyatakan perbandingan antara jumlah mol sebuah unsur dengan jumlah total seluruh unsur dalam satu larutan. Fraksi mol tidak memiliki satuan dan dinotasikan dengan X dan Fraksi mol total selalu bernilai satu. Rumus untuk mencari fraksi mol ialah :
X_A =

nA / nA + nB

dan X_B =

nB / nA + nB

Dimana :

nA adalah mol zat pelarut
nB yaitu mol zat terlarut
X_A yaitu fraksi mol pelarut
X_B yaitu fraksi mol terlarut

Jumlah fraksi mol zat terlarut dan pelarut yakni 1 :
X_A + X_B = 1

atau secara biasa fraksi mol mampu mampu dirumuskan sebagai berikut :
X =

mol zat pelarut / mol pelarut

n₁ / n₁ + n₂

Hubungan fraksi mol dengan molalitas :
m =

fraksi mol zat pelarut x 1000 / (1 – fraksi mol terlarut) x M_r pelarut

Contoh Soal Fraksi Mol Soal No.1 Suatu larutan berisikan 2 mol zat A, 3 mol zat B, dan 5 mol zat C. Hitung fraksi mol masing-masing zat

Pembahasan X_A =

nA / nA + nB +nC

2 / (2 + 3 + 5)

= 0.2
X_B =

nB / nA + nB +nC

3 / (2 + 3 + 5)

= 0.3
X_C =

nC / nA + nB +nC

5 / (2 + 3 + 5)

= 0.5
X_A + X_B + X_C = 1

Soal No.2 Hitunglah nilai fraksi mol etanol (Mr = 46) dalam larutan etanol 46% massa ? Pembahasan Dalam soal dibilang larutan etanol 46% massa, artinya bahwa setiap 100 gram larutan mengandungung :
massa etanol =

46 / 100

x 100 = 46 gram
massa air = 100 – 46 = 54 gram
Jumlah mol etanol (nB) =

46 / 46

= 1 mol
Jumlah mol air (nA) =

54 / 18

= 3 mol
Kaprikornus fraksi mol etanol (X_B) :
X_B =

nB / nA + nB

1 / 1 + 3

= 0,25

Soal No.3 Fraksi mol natrium hidroksida sebuah larutan NaOH dalam air adalah 0,05. Berapa fraksi mol air ? Pembahasan X_NaOH + X_Air = 1
0,05 + X_Air = 1
X_Air = 1 – 0,05
X_Air = 0,95 Soal No.4 Berapa kadar (dalam %) glukosa (C₆H₁₂O₆)dalam suatu larutan, kalau diketahui fraksi mol glukosa dalam larutan itu sama dengan 0,2 (Mr glukosa = 98) ? Pembahasan X_glukosa + X_air = 1
0,2 + X_air = 1
X_air = 1 – 0,2 = 0,8

Perbandingan jumlah kedua mol

fraksi mol glukosa / fraksi mol air

0,2 / 0,8

1 / 4

massa glukosa / massa air

fraksi mol glukosa x Mr glukosa / fraksi mol air x Mr air

1 x 98 / 4 x 18

98 / 72

Kadar (%) glukosa =

98 / 72

x 100% = 0,58%
Jadi kadar glukosa dalam larutan sebesar 0,58%

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