A solution of two components containing n1 moles of the 1st component and n2 moles of the 2nd compon

Q: Why can we just assume the total moles to be 1? Doesn't that change the problem?

Concentration is an intensive property, meaning it doesn't depend on the amount of solution you have. Whether you have 1 drop or 1 ocean of this solution, the relationship between molarity and mole fraction remains identical. Assuming 1 mole just makes the math easier!

Q: I tried solving this without assuming 1 mole, and I got stuck in a massive algebraic mess. Is it wrong?

It's not wrong, but it's highly inefficient. You would have to express $n_1$ as $n_2(1-\chi_2)/\chi_2$ and substitute it everywhere. The 'basis assumption' is a standard JEE trick to bypass this exact algebraic nightmare.

Q: Why did we divide the volume by 1000?

The density is given in $\text{g mL}^{-1}$. When you divide mass (in grams) by density, the resulting volume is in milliliters (mL). However, the strict definition of molarity requires the volume to be in Liters (L). Hence, the division by 1000.

Q: What if the question asked for molality instead of molarity?

For molality, you wouldn't even need the density! You would just take the mass of the solvent ($n_1 M_1$), convert it to kg, and divide the moles of solute ($n_2$) by it. Density is the specific bridge needed to cross from mass-based units to volume-based units.

Q: Why is the mass of the solution $n_1 M_1 + n_2 M_2$?

The total mass of any mixture is simply the sum of the masses of its individual components. Since Mass = Moles $\times$ Molar Mass, the mass of the solvent is $n_1 M_1$ and the mass of the solute is $n_2 M_2$.

Question

A solution of two components containing $n_1$ moles of the 1st component and $n_2$ moles of the 2nd component is prepared. $M_1$ and $M_2$ are the molecular weights of component 1 and 2 respectively. If $d$ is the density of the solution in $	ext{g mL}^{-1}$, $C_2$ is the molarity and $\chi_2$ is the mole-fraction of the 2nd component, then $C_2$ can be expressed as

Options:

(A) $C_2 = \frac{1000 \chi_2}{M_1 + \chi_2(M_2 - M_1)}$

(B) $C_2 = \frac{d \chi_2}{M_2 + \chi_2(M_2 - M_1)}$

(C) $C_2 = \frac{1000 d \chi_2}{M_1 + \chi_2(M_2 - M_1)}$

(D) $C_2 = \frac{d \chi_1}{M_2 + \chi_2(M_2 - M_1)}$

Sigma Percentile · Accepted Answer

### The Art of Concentration Conversions

Imagine you are standing in a chemistry lab, staring at a beaker filled with a perfectly mixed solution. You know the mole fraction of the solute, you know the molar masses of the components, and you know the density of the liquid. But suddenly, you need the molarity. How do you bridge the gap between these seemingly disconnected worlds of moles, mass, and volume?

This problem is a classic test of your algebraic agility and conceptual clarity. Many students look at a problem filled entirely with variables—$n_1$, $n_2$, $M_1$, $M_2$, $d$, $\chi_2$—and immediately feel intimidated. But I promise you, once you understand the physical meaning behind the math, this derivation becomes a beautiful, logical dance.

### Setting the Stage: The Variables

Let's break down what we have. We are dealing with a binary solution—meaning it has exactly two components.

Component 1 is our solvent. It has $n_1$ moles and a molar mass of $M_1$.
Component 2 is our solute. It has $n_2$ moles and a molar mass of $M_2$.

Our ultimate goal is to find the molarity of the solute, denoted as $C_2$. By definition, molarity is the number of moles of solute divided by the volume of the solution in Liters:

$$C_2 = \frac{n_2}{V_{	ext{solution}} 	ext{ (in L)}}$$

We are also given the mole fraction of the solute, $\chi_2$, which is defined as:

$$\chi_2 = \frac{n_2}{n_1 + n_2}$$

### The "One Mole" Masterstroke

Here is where the magic happens. If you try to solve this by keeping both $n_1$ and $n_2$ in your equations, you will end up in a massive algebraic swamp. You'd have to express $n_1$ in terms of $n_2$ and $\chi_2$, substitute it into the mass equation, take common denominators, and pray you don't drop a minus sign.

But concentration is an **intensive property**. This means the molarity of a solution is exactly the same whether you have a tiny test tube of it or a massive swimming pool of it. Because the amount doesn't matter, we can assume any convenient amount to make our math easier. This is called assuming a "basis."

Let's make the smartest assumption possible: **Assume the total number of moles in our sample is exactly 1.**

$$n_1 + n_2 = 1$$

Look at what this does to our mole fraction equation! If the denominator $n_1 + n_2$ is 1, then the moles of the solute ($n_2$) is simply equal to its mole fraction ($\chi_2$).

$$n_2 = \chi_2$$

And consequently, the moles of the solvent ($n_1$) is simply $1 - \chi_2$.

$$n_1 = 1 - \chi_2$$

Suddenly, the variables $n_1$ and $n_2$ have vanished, replaced entirely by the mole fraction $\chi_2$. This is a massive victory.

### From Moles to Mass

Now that we know exactly how many moles of each component we have in our 1-mole sample, let's find the total mass of this sample. The total mass is just the mass of the solvent plus the mass of the solute.

$$	ext{Total Mass} = (	ext{Moles of 1} 	imes M_1) + (	ext{Moles of 2} 	imes M_2)$$

Substituting our new expressions for moles:

$$	ext{Total Mass} = (1 - \chi_2)M_1 + \chi_2 M_2$$

Let's expand and rearrange this to make it look cleaner:

$$	ext{Total Mass} = M_1 - \chi_2 M_1 + \chi_2 M_2$$
$$	ext{Total Mass} = M_1 + \chi_2(M_2 - M_1)$$

This expression represents the mass (in grams) of exactly 1 mole of our solution.

### The Density Bridge: Mass to Volume

We have the mass, but molarity requires volume. This is where density ($d$) steps in. Density is the bridge that connects the world of mass to the world of volume.

$$	ext{Volume} = \frac{	ext{Mass}}{	ext{Density}}$$

Substituting our mass expression:

$$V_{	ext{mL}} = \frac{M_1 + \chi_2(M_2 - M_1)}{d}$$

**Warning!** Don't make a silly mistake here. The density was given in $	ext{g mL}^{-1}$. Because we divided grams by $	ext{g mL}^{-1}$, our resulting volume is strictly in milliliters (mL).

Molarity demands that the volume be in Liters. To convert mL to Liters, we must divide by 1000.

$$V_{	ext{L}} = \frac{M_1 + \chi_2(M_2 - M_1)}{1000 d}$$

### The Final Assembly

We are finally ready to assemble our molarity equation.

$$C_2 = \frac{	ext{Moles of solute}}{V_{	ext{L}}}$$

Remember our brilliant assumption? The moles of solute ($n_2$) is simply $\chi_2$. Let's plug everything in:

$$C_2 = \frac{\chi_2}{\frac{M_1 + \chi_2(M_2 - M_1)}{1000 d}}$$

When you divide by a fraction, you multiply by its reciprocal. The $1000 d$ flips to the numerator, giving us our final, elegant expression:

**$$C_2 = \frac{1000 d \chi_2}{M_1 + \chi_2(M_2 - M_1)}$$**

And there it is. By understanding the physical meaning of intensive properties and using the "basis assumption" trick, we bypassed a mountain of algebra and arrived straight at the correct answer. Keep this technique in your arsenal; it will save you precious minutes in the exam hall!

Select Answer:

Visualized Solution

System Setup

Core Definitions

The 1-Mole Assumption

Moles in terms of Mole Fraction

Mass of the Solution

Volume of the Solution

Unit Conversion

Final Molarity Expression

Conclusion

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

The Art of Concentration Conversions

Setting the Stage

The "One Mole" Masterstroke

From Moles to Mass

The Density Bridge

The Final Assembly

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