Animated Solution for Chemistry - Basic Concepts in Chemistry: The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its molality. Density of this solution at 298 K is 2.0 g cm${}^{-3}$. The ratio of the molecular weights of the solute and solvent, $\left(\frac{M{W}_{\text{solute}}}{M{W}_{\text{solvent}}}\right)$, is

Every great chemistry problem starts with a careful reading of the given parameters. In this scenario, we are handed a solution where the mole fraction of the solute is $X_{\text{solute}}=0.1$. Because a solution is entirely made up of solute and solvent, their mole fractions must sum to exactly $1$. This immediately tells us that the mole fraction of the solvent is $X_{\text{solvent}}=0.9$.

But the real magic of this problem lies in a very special, highly unusual constraint: the molarity ($M$) of the solution is exactly equal to its molality ($m$). This is not a general rule; it is a specific condition that we must exploit.

To bridge the gap between these concentration terms, we are also given the density of the solution, $d=2.0{\text{ g/cm}}^3$ (which is equivalent to $2.0\text{ g/mL}$). Our ultimate goal is to find the ratio of the molecular weights of the solute and the solvent, $\left(\frac{M{W}_{\text{solute}}}{M{W}_{\text{solvent}}}\right)$.

To understand why $M=m$ is such a powerful statement, we need to look at their mathematical definitions. Molarity is defined as the number of moles of solute divided by the volume of the solution in liters:

Molality, on the other hand, is the number of moles of solute divided by the mass of the solvent in kilograms:

Notice the beautiful symmetry here? Both terms share the exact same numerator: $n_{\text{solute}}$. When we equate the two expressions because $M=m$, the moles of solute cancel out completely!

This leaves us with a profound, direct relationship: the numerical value of the solution's volume in liters is exactly equal to the numerical value of the solvent's mass in kilograms.

Working with liters and kilograms can be slightly cumbersome when dealing with laboratory-scale densities given in grams and milliliters. Let's make our lives easier by converting these units.

If we multiply both sides of our equality by $1000$, the relationship holds perfectly true for smaller units. The volume of the solution in milliliters becomes numerically equal to the mass of the solvent in grams.

To keep our algebra clean, let's assign this common numerical value a variable, say $V$. So, the volume of the solution is $V\text{ mL}$, and the mass of the solvent is $V\text{ grams}$.

Now, we need to find the total mass of the solution. This is where the density comes into play. Density is mass divided by volume, which means mass is volume multiplied by density.

Substituting our volume $V$ and the given density of $2.0\text{ g/mL}$, we find that the total mass of the solution is $2V\text{ grams}$.

Think about the physical reality of a solution. It is simply a mixture. Therefore, the total mass of the solution must be the sum of the mass of the solute and the mass of the solvent. This is the law of conservation of mass in action.

We already know that the total mass of the solution is $2V$, and the mass of the solvent is $V$. Let's plug these values into our conservation equation.

Solving for the mass of the solute, we get a massive breakthrough: $W_{\text{solute}}=V\text{ grams}$.

This means that the mass of the solute is exactly equal to the mass of the solvent!

Imagine a perfectly balanced scale. This incredible symmetry is the key that unlocks the final answer.

We have a relationship between masses, but we need a relationship between molecular weights. How do we connect them? We use the fundamental mole concept: mass equals the number of moles multiplied by the molecular weight ($W=n\times MW$).

Let's substitute this into our mass equality:

We need the ratio of the molecular weights, so let's rearrange the equation to isolate that ratio on one side.

Notice that the ratio of the molecular weights is the inverse of the ratio of their moles. But wait, what is the ratio of their moles?

Remember the mole fractions from the very beginning? The ratio of the number of moles of two components in a mixture is exactly equal to the ratio of their mole fractions.

Since the ratio of the moles of solute to solvent is $1/9$, the inverse ratio (solvent to solute) must be $9/1$, or simply $9$.

And there we have it! By carefully unpacking the definitions of molarity and molality, and trusting the algebra of mass conservation, we arrived at a beautifully elegant integer answer. The final ratio is 9.