Tentukan titik didih dan titik beku larutan urea 30 g/500 g air adalah pertanyaan klasik dalam mempelajari sifat koligatif larutan. Konsep ini menjelaskan fenomena menarik di mana keberadaan zat terlarut yang tidak mudah menguap dapat mengubah titik didih dan titik beku pelarutnya. Pengetahuan ini bukan hanya teori belaka, melainkan aplikasinya dapat kita temui dalam kehidupan sehari-hari, dari proses pembuatan es krim hingga penggunaan antibeku pada radiator mobil.
Untuk menjawab soal tersebut, diperlukan pemahaman tentang larutan nonelektrolit seperti urea, serta langkah-langkah sistematis dalam perhitungan. Dimulai dari menghitung molalitas larutan, kemudian menggunakan tetapan kenaikan titik didih (Kb) dan penurunan titik beku (Kf) air, kita dapat menemukan perubahan suhu yang terjadi dan menentukan titik didih serta titik beku akhir dari larutan.
Sifat Koligatif Larutan: The Lowdown
Alright, let’s break this down. In the world of chemistry, when you dissolve stuff in a liquid, you’re not just making a mixture; you’re actually changing the liquid’s physical properties in a pretty predictable way. These changes are called sifat koligatif, or colligative properties. The key thing here is that it doesn’t matter
-what* you dissolve, but
-how many particles* you dissolve.
It’s all about the particle count, mate.
There are four main ones you need to know: penurunan tekanan uap (vapour pressure lowering), kenaikan titik didih (boiling point elevation), penurunan titik beku (freezing point depression), and tekanan osmotik (osmotic pressure). For our task today, we’re zoning in on the boiling and freezing points.
Larutan Elektrolit versus Nonelektrolit
This particle count business is where the type of solute becomes a massive deal. If you dissolve a nonelektrolit like sugar or urea, the molecules stay as they are. One formula unit equals one particle. But if you dissolve an elektrolit like salt (NaCl), it splits into ions (Na⁺ and Cl⁻). So, one formula unit gives you
-two* particles, which has a much bigger effect on those colligative properties.
That’s why we need a fancy factor called the van’t Hoff factor (i) for electrolytes, but for nonelektrolits like our urea, it’s just 1.
Contoh Nyata dalam Kehidupan Sehari-hari
You see this science in action all the time. In winter, they scatter salt on icy roads. The salt dissolves in the thin layer of water on the ice, forming a solution that freezes at a much lower temperature than 0°C, causing the ice to melt. That’s penurunan titik beku. On the flip side, adding salt to water when cooking pasta actually raises the water’s boiling point slightly (not by loads, but it does), so your pasta cooks in water that’s a tad over 100°C.
That’s kenaikan titik didih.
Membongkar Soal: Data dan Persiapan: Tentukan Titik Didih Dan Titik Beku Larutan Urea 30 g/500 g Air
So, our mission is: “Tentukan titik didih dan titik beku larutan urea 30 g/500 g air.” Sounds simple, but we need to be meticulous. Before we crunch any numbers, we have to identify all the bits and bobs given and make sure everything is in the right units. Mass needs to be in grams or kilograms, and we’ll need the Mr of urea.
Let’s get it all laid out properly.
Identifikasi Data dari Soal, Tentukan titik didih dan titik beku larutan urea 30 g/500 g air
Here’s everything the problem gives us, stripped down and clear. The key is to spot what’s what: the solute, the solvent, their masses, and what we’re asked to find.
| Komponen | Nilai | Satuan | Keterangan |
|---|---|---|---|
| Massa Urea | 30 | gram | Zat terlarut (solute) |
| Massa Air | 500 | gram | Pelarut (solvent) |
| Pelarut | Air | – | Dengan Kb dan Kf tertentu |
| Ditanya | Titik Didih & Titik Beku Larutan | °C | Hasil akhir yang dicari |
Langkah Awal dan Konversi Satuan
The first proper step is always to get the solvent mass into kilograms. Chemistry loves its SI units. Five hundred grams is 0.5 kg. Easy. Next, we absolutely must calculate the molar mass (Mr) of urea, CO(NH₂)₂.
Without that, we can’t find out how many moles we have, and the whole calculation falls apart.
Menghitung Molalitas: The Heart of the Matter
Molalitas (m) is the concentration star of the show for colligative properties. It’s defined as the number of moles of solute per kilogram of solvent. It’s perfect for this because it doesn’t change with temperature, unlike molarity. So, let’s figure out how concentrated our urea solution is.
Massa Molekul Relatif (Mr) Urea
Urea’s formula is CO(NH₂)₂, which is often written as CH₄N₂O. Let’s sum up the atomic masses: C=12, O=16, N=14, H=1. So, Mr = 12 + 16 + (2*14) + (4*1) = 60 g/mol. Sorted.
Prosedur Perhitungan Molalitas
First, find the moles of urea: n = massa / Mr = 30 g / 60 g/mol = 0.5 mol. We already converted the solvent: 500 g = 0.5 kg. Now, plug it into the molality formula:
m = (massa / Mr) / kg pelarut = (30 / 60) / 0.5 = 0.5 / 0.5 = 1.0 molal
So, our solution is a 1.0 molal urea solution. Nice and round.
Kenaikan Titik Didih Larutan (ΔTb)
Right, now for the fun bit. Because we’ve added a solute, the solution will boil at a higher temperature than pure water. The increase, ΔTb, is directly proportional to the molality. The proportionality constant is Kb, the ebullioscopic constant, which is unique to each solvent.
Rumus dan Tetapan Kenaikan Titik Didih
For a nonelektrolit, the formula is dead simple:
ΔTb = Kb × m
For water, Kb is well-known: 0.52 °C/m or °C kg mol⁻¹. This means a 1 molal solution of a nonelectrolyte in water will boil 0.52°C higher than pure water.
Perhitungan Titik Didih Larutan Urea
We have m = 1.0 and Kb = 0.52. So, ΔTb = 0.52 °C/m × 1.0 m = 0.52 °C. The normal boiling point of pure water is 100°C. Therefore, the boiling point of our solution is 100°C + 0.52°C = 100.52°C.
Penurunan Titik Beku Larutan (ΔTf)
Similarly, the presence of a solute lowers the freezing point. The decrease, ΔTf, is also directly proportional to molality, with its own constant, Kf, the cryoscopic constant.
Rumus dan Tetapan Penurunan Titik Beku
The formula is the mirror image of the boiling point one:
ΔTf = Kf × m
For water, Kf is 1.86 °C/m. A 1 molal nonelectrolyte solution will freeze 1.86°C lower than pure water.
Perhitungan Titik Beku Larutan Urea
Using our molality of 1.0: ΔTf = 1.86 °C/m × 1.0 m = 1.86 °C. The normal freezing point of water is 0°C. So, the freezing point of our solution is 0°C – 1.86°C = -1.86°C.
Ringkasan Hasil dan Analisis
Let’s pull all the results together in one place to see the full picture. Comparing these numbers to pure water really shows the effect of dissolving just 30 grams of urea in half a kilo of water.
Tabel Ringkasan Hasil Perhitungan
| Parameter | Nilai | Satuan |
|---|---|---|
| Molalitas (m) | 1.00 | mol/kg |
| Kenaikan Titik Didih (ΔTb) | 0.52 | °C |
| Titik Didih Larutan | 100.52 | °C |
| Penurunan Titik Beku (ΔTf) | 1.86 | °C |
| Titik Beku Larutan | -1.86 | °C |
Interpretasi Hasil Numerik
The numbers tell a clear story. A 1 molal solution isn’t even that concentrated, but it has a measurable effect.
The solution boils at 100.52°C and freezes at -1.86°C. This means you’d have to heat it slightly more to get it to boil, and you’d have to cool it well below 0°C to get it to solidify. This demonstrates how dissolving a non-volatile solute stabilises the liquid phase, making it harder to escape as vapour (boiling) and harder to form a solid lattice (freezing).
Eksplorasi Variasi Soal dan Aplikasi
To really get our heads around how these factors work, let’s tweak the original problem in a couple of ways. Seeing how the final answers shift when we change one variable at a time is proper enlightening.
Variasi Soal 1: Massa Urea Digandakan
Source: amazonaws.com
Scenario: What if we dissolve 60 g of urea in 500 g air? Molality becomes: n = 60/60 = 1 mol; kg solvent = 0.5 kg; m = 1/0.5 = 2.0 molal. Then, ΔTb = 0.52
– 2 = 1.04°C (Tb = 101.04°C). ΔTf = 1.86
– 2 = 3.72°C (Tf = -3.72°C). Doubling the solute mass doubles the molality, which directly doubles the ΔTb and ΔTf.
Variasi Soal 2: Pelarut Diganti (Misal, Benzena)
Scenario: 30 g urea in 500 g benzena (C₆H₆). Kb benzena = 2.53 °C/m, Kf = 5.12 °C/m, Tb pure = 80.1°C, Tf pure = 5.5°C. Molality stays 1.0 (same masses, same Mr). But now: ΔTb = 2.53
– 1 = 2.53°C (Tb larutan = 80.1 + 2.53 = 82.63°C). ΔTf = 5.12
– 1 = 5.12°C (Tf larutan = 5.5 – 5.12 = 0.38°C).
The change is much larger because benzena’s constants are bigger.
Prosedur Perbandingan Ketiga Skenario
- Skenario Awal (1 m in water): Tb = 100.52°C, Tf = -1.86°C. The baseline effect.
- Variasi 1 (2 m in water): Tb = 101.04°C, Tf = -3.72°C. Shows the direct, linear proportionality to molality. More solute = bigger change.
- Variasi 2 (1 m in benzene): Tb = 82.63°C, Tf = 0.38°C. Dramatically highlights that the magnitude of change depends crucially on the solvent’s inherent properties (Kb and Kf). The same molality causes a nearly 5x greater freezing point depression in benzene compared to water.
Kesimpulan
Dari perhitungan yang telah dilakukan, terlihat jelas pengaruh zat terlarut terhadap sifat fisik pelarut. Larutan urea 30 gram dalam 500 gram air memiliki titik didih yang sedikit lebih tinggi dan titik beku yang lebih rendah dibandingkan air murni. Perubahan yang relatif kecil ini menguatkan konsep sifat koligatif, di mana efeknya bergantung pada jumlah partikel zat terlarut. Pemahaman ini membuka wawasan tentang bagaimana ilmu kimia menjelaskan berbagai fenomena di sekitar kita dengan angka dan logika yang sederhana namun powerful.
FAQ Terpadu
Mengapa urea termasuk larutan nonelektrolit dalam perhitungan ini?
Urea (CO(NH₂)₂) merupakan senyawa kovalen yang ketika dilarutkan dalam air tidak terionisasi menjadi partikel bermuatan (ion). Ia tetap berupa molekul utuh. Karena sifat koligatif bergantung pada jumlah partikel zat terlarut, dan urea tidak menghasilkan ion, maka ia diperlakukan sebagai nonelektrolit dengan faktor van’t Hoff (i) = 1.
Bagaimana jika air sebagai pelarut diganti dengan pelarut lain seperti benzena?
Hasil titik didih dan titik beku akan sangat berbeda karena setiap pelarut memiliki tetapan Kb dan Kf yang unik. Misalnya, benzena memiliki Kb dan Kf yang tidak sama dengan air. Selain itu, titik didih dan titik beku normal pelarut murninya juga berbeda, sehingga perhitungan harus dimulai dari nilai tetapan dan titik referensi yang baru.
Apakah massa jenis air perlu diperhitungkan dalam konversi 500 g air menjadi volume?
Untuk perhitungan molalitas (m), yang dibutuhkan adalah massa pelarut dalam kilogram, bukan volumenya. Karena soal sudah memberikan massa air sebesar 500 gram (atau 0.5 kg), konversi massa jenis tidak diperlukan sama sekali. Molalitas bergantung pada rasio mol zat terlarut terhadap massa pelarut, bukan volume larutan.
Mengapa hasil perhitungan titik didih larutan hanya naik sedikit di atas 100°C?
Kenaikan titik didih adalah sifat koligatif yang besarnya proporsional dengan molalitas. Larutan encer seperti dalam soal (30 g urea dalam 500 g air) memiliki molalitas yang rendah, sehingga kenaikan suhunya (ΔTb) juga kecil, biasanya hanya dalam orde beberapa puluh derajat Celsius. Hal ini menunjukkan bahwa efek sifat koligatif pada larutan encer memang tidak dramatis, tetapi terukur.
