Q.Calculate the mass of urea ($NH_2CONH_2$) required in making 2.5 kg of 0.25 molal aqueous solution.

Concept: Molality Calculation

Reasoning:

1. The molar mass of urea ($NH_2CONH_2$) is calculated as $2(14) + 4(1) + 12 + 16 = 60 \text{ g/mol}$.
2. Molality ($m$) is defined as moles of solute per kilogram of solvent. For a 0.25 molal solution, the moles of urea ($n_{\text{urea}}$) relate to the mass of solvent ($W_{\text{solvent}}$ in kg) as $n_{\text{urea}} = 0.25 \times W_{\text{solvent}}$.
3. The total mass of the solution ($2.5 \text{ kg}$) is the sum of the mass of urea ($n_{\text{urea}} \times 0.060 \text{ kg/mol}$) and the mass of solvent ($W_{\text{solvent}}$).

$$2.5 = W_{\text{solvent}} (0.015 + 1) = 1.015 \times W_{\text{solvent}}$$ …

Molality is a measure of the concentration of a solute in a solution, defined as the number of moles of solute per kilogram of solvent. It is particularly useful because, unlike molarity, it does not change with temperature, as both moles and mass are temperature-independent quantities.

In this problem, we are given the molality of the solution and the total mass of the solution, not just the solvent. This is a common point of confusion. We need to determine the mass of urea (solute) required. To do this, we will set up equations based on the definition of molality and the total mass of the solution, then solve for the unknown mass of urea.

Here is the step-by-step calculation:

1. Calculate the molar mass of urea ($NH_2CONH_2$).

   The molar mass is the sum of the atomic masses of all atoms in the molecule.

   • Nitrogen (N): $2 \times 14.01 \text{ g/mol} = 28.02 \text{ g/mol}$
   • Hydrogen (H): $4 \times 1.008 \text{ g/mol} = 4.032 \text{ g/mol}$
   • Carbon (C): $1 \times 12.01 \text{ g/mol} = 12.01 \text{ g/mol}$
   • Oxygen (O): $1 \times 16.00 \text{ g/mol} = 16.00 \text{ g/mol}$

   Molar mass of urea ($M_{urea}$) $= 28.02 + 4.032 + 12.01 + 16.00 = 60.062 \text{ g/mol}$.

   We will use $60.06 \text{ g/mol}$ for our calculations.

2. Define molality and set up initial relationships.

   Molality ($m$) is given by:

   $m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}$

   Let $n_{urea}$ be the moles of urea, $M_{urea}$ be the mass of urea (in grams), and $M_{solvent}$ be the mass of solvent (water, in grams).

   From the definition of molality, we have:

The moles of urea can also be expressed as:

Substituting the expression for $n_{urea}$ into the molality equation:

Rearranging this equation to express $M_{solvent}$ in terms of $M_{urea}$:

3. Use the total mass of the solution to solve for the mass of urea. …

Method: Molality Formula Method

Molality ($m$) is defined as the number of moles of solute per kilogram of solvent.

Step-by-step solution

Step 1: Identify given data

• Molality, $m = 0.25 \, \text{mol/kg}$
• Mass of solution = $2.5 \, \text{kg}$
• Solute = urea ($NH_2CONH_2$)

Step 2: Understand the relationship

In a solution:

Let mass of solute = $w$ grams.

Then mass of solvent (in kg) = $\frac{2.5 \, \text{kg} - w \, \text{kg}}{1}$ — but careful: $w$ is in grams, so convert.

Step 3: Express molality equation

Moles of urea = $\frac{w}{\text{molar mass of urea}}$

Molar mass of urea:

• C = 12, O = 16, N = 14, H = 1
• $12 + 16 + 2(14) + 4(1) = 60 \, \text{g/mol}$

So:

Mass of solvent (in kg) = $\frac{2.5 \times 1000 - w}{1000} = \frac{2500 - w}{1000}$

Step 4: Apply molality formula

Step 5: Solve for $w$

Cross-multiply: …

🧠 First, the concept (why it matters)

Molality ($m$) is moles of solute per kg of solvent, not per kg of solution.

This is the single most important distinction.

Formula:

In this problem:

• $m = 0.25 \, \text{mol/kg}$
• Total mass of solution = 2.5 kg
• Solute = urea ($NH_2CONH_2$), molar mass = $60 \, \text{g/mol}$

✗ Common Mistake #1: Confusing solvent mass with solution mass

What students do:

They take 2.5 kg as the mass of solvent and directly plug into:

Then mass of urea = $0.625 \times 60 = 37.5 \, \text{g}$ — wrong.

Why it’s wrong:

2.5 kg is the total solution mass (solute + solvent). The solvent mass is less than 2.5 kg.

How to avoid:

Always read carefully: “2.5 kg of 0.25 molal aqueous solution” means the solution weighs 2.5 kg.

Let solvent mass = $W$ kg. Then:

Mass of urea = $0.25 \times W \times 60$ grams = $15W$ grams.

Total solution mass = solvent mass + solute mass:

Solve for $W$, then find urea mass.

✗ Common Mistake #2: Using molarity formula instead of molality

What students do:

They use $M = \frac{\text{moles}}{\text{volume in L}}$ and assume 2.5 kg ≈ 2.5 L (for water).

Then moles = $0.25 \times 2.5 = 0.625$, same wrong answer.

Why it’s wrong:

Molality uses mass of solvent, not volume of solution. Also, density of solution ≠ 1 g/mL always.

How to avoid:

Memorise:

• Molarity (M) = moles / volume of solution (L)
• Molality (m) = moles / mass of solvent (kg) Write the correct formula at the start of every problem.

✗ Common Mistake #3: Forgetting to convert grams to kg

What students do:

They solve correctly but give urea mass in kg instead of grams, or vice versa.

Example:

If they get 0.0375 kg, they might write 37.5 g (correct) but sometimes write 0.0375 g (wrong).

How to avoid: …