The molality of a urea solution in which 0.0100 g of urea, [(NH2)2CO] is added to 0.3000 dm^3 of wat

Q: Why did we use the density of water as 1 g/mL?

At standard temperature and pressure (STP), the density of liquid water is very close to $1 \text{ g/mL}$ (or $1 \text{ kg/L}$). This is a standard assumption unless a different density is explicitly provided.

Q: Why didn't we subtract the mass of urea from the total mass to get the solvent mass?

The question states urea is added *to* $0.3000 \text{ dm}^3$ of water. This means the water itself is the solvent, so its mass is exactly the mass of the solvent. We don't need to subtract anything.

Q: What is the difference between molality and molarity?

Molality (m) is moles of solute per kilogram of *solvent*, whereas molarity (M) is moles of solute per liter of *solution*. Molality is temperature-independent!

Q: How do I quickly convert $\text{dm}^3$ to mL?

Remember that $1 \text{ dm}^3$ is exactly equal to $1 \text{ Liter}$. Since $1 \text{ L} = 1000 \text{ mL}$, $0.3000 \text{ dm}^3$ is simply $300 \text{ mL}$.

Q: Is the molar mass of urea always 60?

Yes! Urea is $\text{NH}_2\text{CONH}_2$. Calculating it: $14(\text{N}) + 2(\text{H}) + 12(\text{C}) + 16(\text{O}) + 14(\text{N}) + 2(\text{H}) = 60 \text{ g/mol}$.

Question

The molality of a urea solution in which $0.0100	ext{ g}$ of urea, $[(	ext{NH}_2)_2	ext{CO}]$ is added to $0.3000	ext{ dm}^3$ of water at STP is

Options:

(A) $5.55 	imes 10^{-4}	ext{ M}$

(B) $33.3	ext{ M}$

(C) $3.33 	imes 10^{-2}	ext{ M}$

(D) $0.555	ext{ M}$

Sigma Percentile · Accepted Answer

Welcome to a classic physical chemistry problem that tests your fundamental understanding of concentration terms! At first glance, this might look like a simple plug-and-chug question, but there are a few subtle traps hidden in the units and the definitions. Let's break it down step by step and truly understand what's happening inside that beaker.

### Visualizing the Mixture

Imagine you are standing in a laboratory. In front of you is a beaker containing exactly $0.3000 	ext{ dm}^3$ of pure water. To this water, you carefully add a tiny pinch of urea—just $0.0100 	ext{ g}$ of it.

Our objective is to determine the **molality** of this newly formed solution. Before we dive into the math, it's crucial to clearly identify our components. The substance being dissolved is the **solute**, which in this case is urea. The substance doing the dissolving is the **solvent**, which is water.

### The Core Principle: Molality

In chemistry, there are many ways to express concentration, but molality is special. Why? Because unlike molarity, which depends on the volume of the solution, molality depends strictly on the mass of the solvent.

The formula for molality ($m$) is defined as:
$$m = \frac{n_{	ext{solute}}}{W_{	ext{solvent}} 	ext{ (in kg)}}$$

Because mass does not change with temperature, molality is a **temperature-independent** concentration term. This makes it incredibly useful for experiments involving boiling point elevation or freezing point depression.

### Crunching the Numbers: Moles and Mass

To use our molality formula, we need two pieces of information: the number of moles of our solute (urea) and the mass of our solvent (water) in kilograms.

**Step 1: Finding the Moles of Urea**
We are given the mass of urea as $0.0100 	ext{ g}$. To convert this to moles, we need its molar mass. The chemical formula for urea is $	ext{NH}_2	ext{CONH}_2$. If we add up the atomic masses ($14 	imes 2$ for Nitrogen, $1 	imes 4$ for Hydrogen, $12$ for Carbon, and $16$ for Oxygen), we get exactly $60 	ext{ g/mol}$.

So, the number of moles of urea is:
$$n_{	ext{urea}} = \frac{0.0100}{60} 	ext{ mol}$$
We will leave it as a fraction for now to avoid rounding errors early in the calculation.

**Step 2: Finding the Mass of Water**
This is where many students make a silly mistake! We are given the volume of water as $0.3000 	ext{ dm}^3$. 
First, let's convert this to a more familiar unit. Remember that $1 	ext{ dm}^3$ is exactly equal to $1 	ext{ Liter}$. Therefore, $0.3000 	ext{ dm}^3$ is $0.3000 	ext{ L}$, or $300 	ext{ mL}$.

Now, we need the mass. At standard temperature and pressure (STP), the density of water is approximately $1 	ext{ g/mL}$. 
$$W_{	ext{water}} = 300 	ext{ mL} 	imes 1 	ext{ g/mL} = 300 	ext{ g}$$
To fit our molality formula, we must convert this to kilograms:
$$W_{	ext{water}} = 0.3 	ext{ kg}$$

*(Note: A common misconception is to subtract the mass of the solute from this value. However, the problem states the urea is added **to** the water, meaning the water itself is the pure solvent. Its mass remains exactly $300 	ext{ g}$.)*

### The Final Result

Now we have everything we need. Let's substitute our values into the master equation:
$$m = \frac{\left(\frac{0.0100}{60}ight)}{0.3}$$

Let's simplify the denominator. Multiplying $60$ by $0.3$ gives us $18$.
$$m = \frac{0.0100}{18}$$

When we perform this final division, we get:
$$m = 0.000555... 	ext{ m}$$
Expressing this in scientific notation, we arrive at our final answer:
$$m = 5.55 	imes 10^{-4} 	ext{ m}$$

This matches option (a) perfectly. You've successfully navigated the unit conversions and applied the core concept of molality. Great job!

Animated Solution for Chemistry - Basic Concepts in Chemistry: The molality of a urea solution in which $0.0100 g$ of urea, $[(NH_{2})_{2} CO]$ is added to $0.3000 dm^{3}$ of water at STP is

Select Answer:

Visualized Solution

Visualizing the Solution

Formula for Molality

Moles of Solute (Urea)

Mass of Solvent (Water)

Substituting the Values

Final Calculation

The Way Forward

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

Visualizing the Mixture

The Core Principle

Crunching the Numbers

The Final Result

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