Fraksi mol zat terlarut
$$\bbox[5px, border: 2px solid red] {\textbf{X}_t = \dfrac{n_t}{n_t + n_p}}$$
Fraksi mol zat pelarut
$$\bbox[5px, border: 2px solid red] {\textbf{X}_p = \dfrac{n_p}{n_t + n_p}}$$
Latihan Soal
Soal 01
Sebanyak 30 gram urea \((\ce{CO(NH2)2})\) dilarutkan dalam 256 gram naftalena \((\ce{C10H8})\). Fraksi mol urea adalah…
(Ar H = 1, C = 12, O = 16, N = 14)
(A) \(\dfrac{1}{5}\)
(B) \(\dfrac{4}{5}\)
(C) \(\dfrac{1}{10}\)
(D) \(\dfrac{1}{12}\)
(E) \(\dfrac{1}{24}\)
Jawaban: A
Zat yang terlarut adalah urea, sedangkan pelarutnya adalah naftalena.
Menghitung jumlah mol urea
Mr \((\ce{CO(NH2)2})\) = 12 + 16 + 28 + 4 = 60
\(n_t = \dfrac{\text{massa urea}}{\text{Mr urea}}\)
\(n_t = \dfrac{30}{60}\)
\(n_t = \dfrac{1}{2} \text{ mol}\)
Menghitung jumlah mol naftalena
Mr \((\ce{C10H8})\) = 120 + 8 = 128
\(n_p = \dfrac{\text{massa naftalena}}{\text{Mr naftalena}}\)
\(n_p = \dfrac{256}{128}\)
\(n_p = 2 \text{ mol}\)
Menghitung fraksi mol zat terlarut (urea)
\(\textbf{X}_t = \dfrac{n_t}{n_t + n_p}\)
\(\textbf{X}_t = \dfrac{\dfrac{1}{2}}{\dfrac{1}{2} + 2}\)
\(\textbf{X}_t = \dfrac{\dfrac{1}{2}}{\dfrac{5}{2}}\)
\(\textbf{X}_t = \dfrac{1}{5}\)
Soal 02
Sebanyak 31 ml glikol \((\ce{C2H6O2})\) yang massa jenisnya 1,5 kg/L dilarutkan dalam 90 ml air. Fraksi mol air adalah…
(A) \(\dfrac{3}{23}\)
(B) \(\dfrac{20}{23}\)
(C) \(\dfrac{3}{25}\)
(D) \(\dfrac{22}{25}\)
(E) \(\dfrac{3}{27}\)
Jawaban: B
Menghitung massa glikol
\(\rho = \dfrac{m}{V}\)
\(1,5 \text{ g/ml} = \dfrac{m}{31 \text{ ml}}\)
\(m = 1,5 \times 31 \text{ gram}\)
\(m = 46,5 \text{ gram}\)
Menghitung jumlah mol glikol (zat terlarut)
Mr \((\ce{C2H6O2})\) = 24 + 6 +32 = 62
\(n_t = \dfrac{\text{massa glikol}}{\text{Mr glikol}}\)
\(n_t = \dfrac{46,5}{62}\)
\(n_t = \dfrac{3}{4} \text{ mol}\)
Menghitung jumlah mol air
Mr \((\ce{H2O})\) = 2 + 16 = 18
\(n_p = \dfrac{\text{massa air}}{\text{air}}\)
\(n_p = \dfrac{90}{18}\)
\(n_p = 5 \text{ mol}\)
Menghitung fraksi mol zat pelarut (air)
\(\textbf{X}_p = \dfrac{n_p}{n_t + n_p}\)
\(\textbf{X}_p = \dfrac{5}{\dfrac{3}{4} + 5}\)
\(\textbf{X}_p = \dfrac{5}{\dfrac{23}{4}}\)
\(\textbf{X}_p = \dfrac{20}{23}\)
Soal 03
Larutan \(\ce{NaOH}\) (Mr = 40) 20% dalam air memiliki fraksi mol…
(A) \(\dfrac{5}{89}\)
(B) \(\dfrac{6}{89}\)
(C) \(\dfrac{7}{89}\)
(D) \(\dfrac{8}{89}\)
(E) \(\dfrac{9}{89}\)
Jawaban: E
Misal
Massa NaOH = 20 gram
Massa air = 80 gram
Menghitung jumlah mol \(\ce{NaOH}\)
\(\text{Jumlah mol NaOH} = \dfrac{\text{massa}}{\text{Mr}}\)
\(\text{Jumlah mol NaOH} = \dfrac{20}{40}\)
\(\text{Jumlah mol NaOH} = \dfrac{1}{2} \text{ mol}\)
Menghitung jumlah mol air
\(\text{Jumlah mol air} = \dfrac{\text{massa}}{\text{Mr}}\)
\(\text{Jumlah mol air} = \dfrac{80}{18}\)
\(\text{Jumlah mol air} = \dfrac{40}{9} \text{ mol}\)
Menghitung fraksi mol NaOH
\(\textbf{X}_t = \dfrac{n_t}{n_t + n_p}\)
\(\textbf{X}_t = \dfrac{\dfrac{1}{2}}{\dfrac{1}{2} + \dfrac{40}{9}}\)
\(\textbf{X}_t = \dfrac{\dfrac{1}{2}}{\dfrac{1}{2} + \dfrac{40}{9}}\)
\(\textbf{X}_t = \dfrac{\dfrac{1}{2}}{\dfrac{89}{18}}\)
\(\textbf{X}_t =\dfrac{1}{2} \times \dfrac{18}{89}\)
\(\textbf{X}_t =\dfrac{9}{89}\)
