Kimia 12 IPA. Tolong dijawab

Vp (volume air) = 2.5 Liter = 2.5 dm³ = 2500 cm³
ρ air = 1 gram/cm³
Tf = -0.74°C
Kf air = 1.86°C/m
Mr t (Mr gula) = 342
mt (massa gula) = . . . . . ?

Penyelesaian
* Mencari masda air
ρ air = m air / V air
1 = mp / 2500
mp = 2500 gram
mp = 2.5 kg

* Mencari penurunan titik beku
Tf = Tf° - ∆Tf
-0.74 = 0 - ∆Tf
∆Tf = 0 + 0.74
∆Tf = 0.74°C

* Mencari molalitas
∆Tf = m . Kf
0.74 = m . 1.86
m = 0.74/1.86
m = 0.398 molal

* Mencari mol gula (nt)
m = nt/mp
0.398 = nt/2.5
nt = 0.398 . 2.5
nt = 0.994 mol

* Mencari masda gula
nt = mt/Mr t
0.994 = mt/342
mt = 0.994 . 342
mt = 340 gram

Jadi, jumlah gula yang harus dilarutkan sebanyam 340 gram (Jawaban E)

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Tingkat : SMA
Kelas : 12
Pelajaran : Kimia
BAB : Sifat Koligatif
Kata Kunci : Titik beku
Kode Kategorisasi : 12.7.1

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