The mole fraction of urea in an aqueous urea solution containing 900 g of water is 0.05. If the dens

Q: Why can't we assume the volume of the solution is just the volume of the water?

When a solute dissolves, it occupies space and changes the overall volume of the mixture. You must use the total mass and the density of the solution to find the true final volume.

Q: How did we get the equation $0.05 = n / (n + 50)$?

Mole fraction is defined as the moles of the component divided by the total moles in the mixture. Here, the total moles equal the moles of urea ($n$) plus the moles of water ($50$).

Q: Why do we need the molar mass of urea if we already found its moles?

We found the moles of urea, but to use density (which is mass/volume), we need the total mass of the solution. The molar mass is required to convert the moles of urea into grams so we can add it to the mass of the water.

Q: Can we solve this by assuming 1 liter of solution from the start?

While assuming a volume works for intensive properties, it's harder here because the given information ($900 \text{ g}$ of water) anchors the absolute amounts. It's much easier to work forward from the given absolute mass of water.

Q: What if the density was given in kg/L instead of g/cm³?

$1.2 \text{ g/cm}^3$ is mathematically exactly equal to $1.2 \text{ kg/L}$. The numerical value remains the same, so the calculation wouldn't change at all!

Question

The mole fraction of urea in an aqueous urea solution containing 900 g of water is 0.05. If the density of the solution is $1.2 	ext{ g cm}^{-3}$, the molarity of urea solution is ___ 
(Given data : Molar masses of urea and water are $60 	ext{ g mol}^{-1}$ and $18 	ext{ g mol}^{-1}$, respectively)

Sigma Percentile · Accepted Answer

### The Interconnected Web of Concentration Terms

In physical chemistry, concentration terms like molarity, molality, and mole fraction are essentially different languages describing the exact same physical reality. The true test of mastery is your ability to translate between these languages seamlessly.

In this problem, we are given the mole fraction of urea and asked to find its molarity. This requires a systematic journey from moles to mass, and finally to volume.

### Decoding the Solvent

Every journey begins with a solid foundation. Here, our foundation is the $900 	ext{ g}$ of water. Before we can use the mole fraction, we need to know exactly how many particles of water we are dealing with.

We calculate the moles of water ($n_{	ext{water}}$) by dividing its given mass by its molar mass ($18 	ext{ g mol}^{-1}$):

$$n_{	ext{water}} = \frac{900}{18} = 50 	ext{ mol}$$

### Unlocking the Solute via Mole Fraction

Now we bring in the mole fraction of urea, $X_{	ext{urea}} = 0.05$. By definition, the mole fraction is the ratio of the moles of the solute to the total moles in the solution.

$$X_{	ext{urea}} = \frac{n_{	ext{urea}}}{n_{	ext{urea}} + n_{	ext{water}}}$$

Substituting our known values, we get:

$$0.05 = \frac{n_{	ext{urea}}}{n_{	ext{urea}} + 50}$$

**A word of caution:** A very common mistake is to write the denominator as just $50$. Remember, the denominator must represent the *total* moles, which includes the solute itself! Solving this linear equation gives us the exact moles of urea:

$$0.05n_{	ext{urea}} + 2.5 = n_{	ext{urea}}$$

$$0.95n_{	ext{urea}} = 2.5 \implies n_{	ext{urea}} = \frac{50}{19} \approx 2.6315 	ext{ mol}$$

### The Bridge from Moles to Mass

To find molarity, we need the volume of the solution. However, the problem only provides the density. Density is the bridge between mass and volume, which means we first need to find the **total mass of the solution**.

The total mass is the sum of the mass of the water and the mass of the urea. We already know the water weighs $900 	ext{ g}$. Let's find the mass of the urea by multiplying its moles by its molar mass ($60 	ext{ g mol}^{-1}$):

$$M_{	ext{urea}} = \frac{50}{19} 	imes 60 \approx 157.89 	ext{ g}$$

Adding this to the water gives us the total mass:

$$M_{	ext{solution}} = 900 + 157.89 = 1057.89 	ext{ g}$$

### Density: The Gateway to Volume

With the total mass in hand, we can finally unlock the volume using the given density of $1.2 	ext{ g cm}^{-3}$ (which is equivalent to $1.2 	ext{ g mL}^{-1}$).

$$V_{	ext{solution}} = \frac{M_{	ext{solution}}}{	ext{Density}} = \frac{1057.89}{1.2} \approx 881.58 	ext{ mL}$$

### The Final Destination: Molarity

We have arrived at the final step. Molarity ($M$) is defined as the moles of solute per liter of solution. We have our moles of urea ($2.6315 	ext{ mol}$) and our volume in liters ($0.88158 	ext{ L}$).

$$M = \frac{2.6315}{0.88158} \approx 2.98 	ext{ mol L}^{-1}$$

By carefully following the logical chain—from mole fraction to moles, to total mass, to volume, and finally to molarity—we've cracked the problem. The final molarity of the urea solution is **$2.98 	ext{ M}$**.

Enter Numerical Value:

Visualized Solution

System Setup: $H_{2} O$ and $N H_{2} C O N H_{2}$

Moles of Solvent ( $n_{water}$ )

Mole Fraction Equation

Solving for $n_{urea}$

Total Mass of Solution ( $M_{solution}$ )

Volume of Solution ( $V_{solution}$ )

Calculating Molarity ( $M$ )

The Logical Flow

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

The Interconnected Web of Concentration Terms

Decoding the Solvent

Unlocking the Solute via Mole Fraction

The Bridge from Moles to Mass

Density

The Final Destination

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Enter Numerical Value:

Visualized Solution

System Setup: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H2​O and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } NH2​CONH2​

Moles of Solvent ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } nwater​)

Mole Fraction Equation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } nurea​

Total Mass of Solution ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Msolution​)

Volume of Solution ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vsolution​)

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The Logical Flow

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

The Interconnected Web of Concentration Terms

Decoding the Solvent

Unlocking the Solute via Mole Fraction

The Bridge from Moles to Mass

Density

The Final Destination

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Solving for $n_{urea}$

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Volume of Solution ( $V_{solution}$ )

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