Animated Solution for Chemistry - Solutions: A $5.25\%$ solution of a substance is isotonic with a $1.5\%$ solution of urea (molar mass $=60{\text{ g mol}}^{-1}$) in the same solvent. If the densities of both the solutions are assumed to be equal to $1.0{\text{ g cm}}^{-3}$, molar mass of the substance will be

The concept of isotonic solutions is a classic and highly scoring topic in physical chemistry. When a problem states that two solutions are isotonic, it is handing you the master key to unlock the entire calculation. Let's dive into the thought process behind solving this elegant problem.

What exactly does it mean for two solutions to be isotonic? In simple terms, it means they exert the exact same osmotic pressure at a given temperature.

Mathematically, we write this as:

We know that osmotic pressure is given by the formula $\pi =CRT$, where $C$ is the molar concentration, $R$ is the universal gas constant, and $T$ is the temperature. Since both solutions are in the same environment, $R$ and $T$ are constant. This leads us to a beautiful simplification: their molar concentrations must be perfectly equal.

Concentration is defined as the number of moles of solute divided by the volume of the solution ($C=\frac{n}{V}$). To find the concentration, we need to deal with the percentages given in the problem: a $5.25\%$ solution of an unknown substance and a $1.5\%$ solution of urea.

Whenever you see percentages like this, make your life easy: assume you have $100\text{ g}$ of each solution. In $100\text{ g}$ of the first solution, you have $5.25\text{ g}$ of the unknown solute. In $100\text{ g}$ of the second solution, you have $1.5\text{ g}$ of urea.

But what about the volume? This is where the density comes into play. The problem states that both solutions have a density of $1.0{\text{ g cm}}^{-3}$ (which is the same as $1.0{\text{ g mL}}^{-1}$). Since Volume = $\frac{\text{Mass}}{\text{Density}}$, the volume for $100\text{ g}$ of either solution is exactly $100\text{ mL}$. Because the densities are identical, the volumes are identical! Let's just call it $V$.

Now, let's set up our master equation by equating the concentrations ($C_1=C_2$):

The volume $V$ beautifully cancels out from both sides, leaving us with just the moles:

Wait, is it really that simple? Yes! Because the volumes are the same, isotonicity here just means the number of moles in $100\text{ g}$ of solution must be equal. Let's substitute the mass and molar mass for each solute. For the unknown substance, the moles are $\frac{5.25}{M}$. For urea (molar mass = $60{\text{ g mol}}^{-1}$), the moles are $\frac{1.5}{60}$.

Now, it's just a matter of simple algebra. Rearranging to solve for $M$:

We know that $\frac{60}{1.5}=40$.

And there we have it! The molar mass of the unknown substance is $210{\text{ g mol}}^{-1}$. By understanding the physical meaning of isotonicity and strategically using a $100\text{ g}$ sample size, the math becomes incredibly clean and straightforward.