Melting Snow: Urea & CaCl₂ Calculation Explained!

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Hey guys! Ever wondered if you could actually melt snow using a simple mixture of chemicals? Well, let's dive into the fascinating world of chemistry and explore whether a mixture of urea and calcium chloride can really do the trick. We’re going to tackle a problem: Can calculations prove that a mixture of 30 grams of urea (CO(NH₂)₂) and 71 grams of CaCl₂ dissolved in 200 ml of water can melt snow at -15°C? This involves some cool concepts, so grab your mental calculators, and let's get started!

Understanding Freezing Point Depression

Before we jump into the calculations, let's quickly understand the key concept behind melting snow with chemicals: freezing point depression. You see, the freezing point of a liquid (like water) is the temperature at which it transitions from a liquid to a solid state (ice). Pure water freezes at 0°C (32°F). However, when you dissolve a solute (like urea or calcium chloride) in a solvent (water), the freezing point of the solution decreases. This means the solution will freeze at a lower temperature than pure water. This freezing point depression is a colligative property, meaning it depends on the number of solute particles in the solution, not the type of particles.

Think of it like this: the solute particles interfere with the water molecules' ability to form the nice, orderly crystal structure of ice. The more solute you add, the more the freezing point goes down. This is why we use salt on icy roads in the winter! The salt dissolves in the water (or melted snow) and lowers the freezing point, preventing the water from refreezing and keeping the roads safer. So, to figure out if our urea and calcium chloride mix can melt snow at -15°C, we need to calculate the freezing point depression caused by these solutes in water.

To really grasp this concept, it's helpful to visualize the water molecules trying to form ice crystals. They want to snuggle up together in a specific arrangement. But when you throw in urea and calcium chloride, these particles act like party crashers, disrupting the ice-forming process. The more party crashers you have, the harder it is for the water molecules to organize, and the lower the temperature needs to be for the water to freeze. So, we're essentially creating a situation where the snow can't freeze at -15°C because the solution's freezing point is even lower. Cool, right? Now, let's crunch some numbers!

Calculating Molar Mass

First things first, we need to calculate the molar masses (Mr) of urea (CO(NH₂)₂) and calcium chloride (CaCl₂). Remember, molar mass is the mass of one mole of a substance, and it's expressed in grams per mole (g/mol). We’ll use the atomic masses (Ar) given in the problem: Ar C=12, H=1, O=16, Ca=40, Cl=35.5.

For urea (CO(NH₂)₂):

  • Mr = Ar C + Ar O + 2 × (Ar N + 2 × Ar H)
  • Mr = 12 + 16 + 2 × (14 + 2 × 1)
  • Mr = 12 + 16 + 2 × (16)
  • Mr = 12 + 16 + 32
  • Mr = 60 g/mol

Next, for calcium chloride (CaCl₂):

  • Mr = Ar Ca + 2 × Ar Cl
  • Mr = 40 + 2 × 35.5
  • Mr = 40 + 71
  • Mr = 111 g/mol

Why are molar masses so important? Well, they allow us to convert the mass of each compound (given in grams) into moles. Moles are like the chemist's counting unit – they tell us how many particles of a substance we have. And remember, freezing point depression depends on the number of particles, so moles are exactly what we need! Think of it like baking a cake: you need to know not just the weight of your ingredients, but also the number of scoops or cups you're using. Molar mass is the key to making that conversion, so we can figure out the 'number of scoops' (moles) for our urea and calcium chloride.

Now that we have the molar masses, we can confidently move on to the next step: calculating the number of moles of each compound in our solution. This is where things start to get really interesting, as we'll see how the different amounts of urea and calcium chloride contribute to the overall freezing point depression. So, let's keep those calculators handy and dive into the next stage of our icy investigation!

Calculating the Number of Moles

Now that we have the molar masses, let's calculate the number of moles of urea and CaCl₂. We'll use the formula:

  • Moles = Mass (g) / Molar mass (g/mol)

For urea (CO(NH₂)₂):

  • Moles = 30 g / 60 g/mol
  • Moles = 0.5 mol

For calcium chloride (CaCl₂):

  • Moles = 71 g / 111 g/mol
  • Moles ≈ 0.64 mol

So, we have 0.5 moles of urea and approximately 0.64 moles of calcium chloride. Remember how we talked about the number of particles being crucial for freezing point depression? Well, these mole values tell us exactly how many 'particle groups' we have for each compound. But here's a little twist: calcium chloride is an ionic compound, which means it dissociates (breaks apart) into ions when dissolved in water. This is super important because each ion acts as an individual particle, further contributing to the freezing point depression!

Imagine you have a bag of marbles (urea) and a bag of LEGO bricks (calcium chloride). Each marble is a single particle, but each LEGO brick can be broken down into multiple pieces. When calcium chloride dissolves, it breaks down into one calcium ion (Ca²⁺) and two chloride ions (Cl⁻). That means one mole of CaCl₂ actually produces three moles of particles in solution (one mole of Ca²⁺ and two moles of Cl⁻). This is called the van't Hoff factor (i), and it's essential for accurately calculating freezing point depression for ionic compounds. So, while we have 0.64 moles of CaCl₂ initially, it effectively contributes 0.64 mol * 3 = 1.92 moles of particles to the solution. This is a significant increase and will have a big impact on the freezing point!

Understanding this dissociation is like unlocking a secret level in our calculation game. We've gone from simply counting molecules to counting the individual ions they produce in solution. This extra layer of complexity is what makes chemistry so fascinating! Now that we know the true number of particles in our solution, we're well-equipped to tackle the final calculation: determining the freezing point depression and seeing if our mixture can indeed melt snow at -15°C. Let’s move on!

Calculating Freezing Point Depression (ΔTf)

Now for the exciting part! We'll calculate the freezing point depression (ΔTf) using the following formula:

  • ΔTf = i * Kf * molality

Where:

  • ΔTf is the freezing point depression (in °C)
  • i is the van't Hoff factor (number of particles the solute dissociates into)
  • Kf is the cryoscopic constant of water (1.86 °C kg/mol)
  • Molality is the moles of solute per kilogram of solvent

First, let's calculate the molality of each solute:

Urea

  • Moles of urea = 0.5 mol
  • Mass of water = 200 ml = 200 g = 0.2 kg (assuming the density of water is 1 g/ml)
  • Molality of urea = 0.5 mol / 0.2 kg
  • Molality of urea = 2.5 mol/kg
  • Since urea is a non-electrolyte (it doesn't dissociate into ions), its van't Hoff factor (i) is 1.

Calcium Chloride

  • Moles of CaCl₂ = 0.64 mol
  • Mass of water = 0.2 kg
  • Molality of CaCl₂ = 0.64 mol / 0.2 kg
  • Molality of CaCl₂ = 3.2 mol/kg
  • As we discussed earlier, CaCl₂ dissociates into 3 ions (Ca²⁺ and 2Cl⁻), so its van't Hoff factor (i) is 3.

Now we can calculate the freezing point depression for each solute:

Urea

  • ΔTf (urea) = 1 * 1.86 °C kg/mol * 2.5 mol/kg
  • ΔTf (urea) = 4.65 °C

Calcium Chloride

  • ΔTf (CaCl₂) = 3 * 1.86 °C kg/mol * 3.2 mol/kg
  • ΔTf (CaCl₂) = 17.856 °C

Finally, let's add the freezing point depressions together to get the total freezing point depression for the solution:

  • ΔTf (total) = ΔTf (urea) + ΔTf (CaCl₂)
  • ΔTf (total) = 4.65 °C + 17.856 °C
  • ΔTf (total) = 22.506 °C

We've reached the heart of our calculation, guys! This 22.506 °C figure represents the total drop in freezing point caused by our mixture of urea and calcium chloride. It's like saying,