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 do we use kg for the solvent instead of liters?

Molality is defined as moles of solute per kilogram of solvent. This makes it independent of temperature, unlike molarity which uses volume and changes as liquids expand or contract with heat.

Q: How did we know the density of water?

The density of water is a standard physical constant. At STP, it is approximately $1 \text{ g/cm}^3$, which perfectly converts to $1 \text{ kg/dm}^3$ or $1 \text{ kg/L}$.

Q: What is the difference between molarity and molality?

Molarity is moles of solute divided by the total volume of the solution in liters. Molality is moles of solute divided by the mass of the solvent in kilograms.

Q: Why is STP mentioned in the question?

STP (Standard Temperature and Pressure) ensures that the density of water is exactly $1 \text{ kg/dm}^3$. If the temperature were significantly higher, the density would be slightly less, changing the mass calculation.

Q: How to calculate the molar mass of urea?

The formula is $\text{NH}_2\text{CONH}_2$. Add the atomic masses: $2 \times 14$ (Nitrogen) + $4 \times 1$ (Hydrogen) + $12$ (Carbon) + $16$ (Oxygen) = $60 \text{ g/mol}$.

Question

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

Options:

(A) $0.555	ext{ m}$

(B) $5.55 	imes 10^{-4}	ext{ m}$

(C) $33.3	ext{ m}$

(D) $3.33 	imes 10^{-2}	ext{ m}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are in a chemistry lab. You have a beaker filled with exactly $0.3000 	ext{ dm}^3$ of pure water. To this, you carefully add a tiny amount, just $0.0100 	ext{ g}$, of urea crystals. Our goal is to find the **molality** of this newly formed solution.

Molality is a very special way to express concentration. Unlike molarity, which depends on the total volume of the solution, molality depends only on the mass of the solvent. This makes it incredibly useful because it doesn't change with temperature!

### The Master Equation
The formula for molality ($m$) is beautifully simple:
$$m = \frac{	ext{Moles of solute}}{	ext{Mass of solvent (in kg)}}$$

To use this formula, we need two pieces of information: the moles of our solute (urea) and the mass of our solvent (water) in kilograms. Let's tackle them one by one.

### Calculating Moles of Urea
First, we need the molar mass of urea. The chemical formula for urea is $	ext{NH}_2	ext{CONH}_2$. By adding up the atomic masses of nitrogen, hydrogen, carbon, and oxygen, we get:
$$	ext{Molar Mass} = 14(2) + 1(4) + 12 + 16 = 60 	ext{ g/mol}$$

Now, we can find the moles of urea by dividing the given mass by the molar mass:
$$n_{	ext{urea}} = \frac{0.0100 	ext{ g}}{60 	ext{ g/mol}}$$

### Converting Volume to Mass for Water
Next, we need the mass of the water. The problem gives us the volume: $0.3000 	ext{ dm}^3$. We know that $1 	ext{ dm}^3$ is exactly equal to $1 	ext{ Liter}$.

At Standard Temperature and Pressure (STP), the density of water is $1 	ext{ g/cm}^3$, which is equivalent to $1 	ext{ kg/L}$ or $1 	ext{ kg/dm}^3$. Therefore, the mass of the water is simply:
$$W_{	ext{water}} = 0.3000 	ext{ dm}^3 	imes 1 	ext{ kg/dm}^3 = 0.3000 	ext{ kg}$$

### Final Calculation
Now, we bring it all together. We substitute our moles of urea and mass of water into the molality formula:
$$m = \frac{\frac{0.0100}{60}}{0.3000}$$

Simplifying the denominator:
$$m = \frac{0.0100}{60 	imes 0.3000} = \frac{0.0100}{18}$$

Calculating the final value:
$$m = 5.55 	imes 10^{-4} 	ext{ m}$$

And there we have it! The molality of our urea solution is $5.55 	imes 10^{-4} 	ext{ m}$.

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

$Visualizing the Solution$

$Formula for Molality$

$Moles of Urea$

$Mass of Water$

$Substituting Values$

$Final Calculation$

$Temperature Independence$

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

Analyzing the Setup Imagine you are in a chemistry lab

The Master Equation

Calculating Moles of Urea First, we need the molar mass of urea

Converting Volume to Mass for Water Next, we need the mass of the water

Final Calculation Now, we bring it all together

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