A \(5.25 \%\) solution of a substance is isotonic with a \(1.5 \%\) solution of urea (molar mass \(=60 \mathrm{~g} \mathrm{~mol}^{-1}\) ) in the same solvent. If the densities of both the solutions are assumed to be equal to \(1.0 \mathrm{~g} \mathrm{~cm}^{-3}\), molar mass of the substance will be: (a) \(115.0 \mathrm{~g} \mathrm{~mol}^{-1}\) (b) \(105.0 \mathrm{~g} \mathrm{~mol}^{-1}\) (c) \(210.0 \mathrm{~g} \mathrm{~mol}^{-1}\) (d) \(90.0 \mathrm{~g} \mathrm{~mol}^{-1}\)

Osmotic pressure is a fundamental concept in chemistry, particularly in the study of solutions and their behaviors. It is the pressure required to stop the flow of a solvent through a semipermeable membrane from a dilute solution into a more concentrated one. This pressure arises because solvents naturally move to equalize concentration differences between two solutions.

In the context of isotonic solutions, osmotic pressure is especially important. Isotonic solutions have the same osmotic pressure, which means when they are separated by a semipermeable membrane there is no net movement of solvent between them. This concept is crucial in biological systems where cells must maintain osmotic balance to function properly. For example, the blood plasma and the cytoplasm within a red blood cell are typically isotonic; this equilibrium prevents the cell from swelling or shrinking, which could cause damage.

When two solutions are isotonic, they have the same solute concentration. To relate this to the exercise, the given concentrations must be expressed as molarities, which can be compared directly to determine whether the solutions will have the same osmotic pressure.

The molarity of a solution is a measure of how many moles of a solute are present in one liter of solution. It's a critical concept when dealing with chemical solutions and reactions. Calculating molarity involves knowing the amount (in grams) of the solute, its molar mass, and the volume of the solution in liters.

To find the molarity, the formula is: \[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]. First, one must convert the mass of the solute into moles by dividing it by the molar mass. Then, divide the number of moles by the volume of the solution in liters.

In the exercise provided, the molarity of a urea solution was obtained by dividing the mass of urea given in the problem by its molar mass and subsequently by the volume of solution in liters. This process is equally applicable to any solute for which you're trying to find molarity, always ensuring that units are consistent for accurate calculations.

Molar mass is the mass of one mole of a substance, and it is a vital molecular-scale characteristic used in chemistry to quantify amounts of substances. The molar mass allows chemists to convert between the mass of a substance and the amount in moles, which is crucial for stoichiometry in chemical equations.

One common method to determine the molar mass of a substance is by rearranging the molarity formula: \[ \text{Molar Mass} = \frac{\text{mass of solute in grams}}{\text{moles of solute}} \]. In this exercise, you would typically know the mass of the solute from the weight percentage of the solution and calculate the moles through molarity calculations. Then, you could solve for the molar mass.

However, using isotonic conditions provides an alternate route. Since the solutions are isotonic, their osmotic pressures and, by extension, their molarities are equal, enabling the comparison of two molarities to find the molar mass of the unknown solute. By setting up an equation based on isotonicity, and knowing the molar mass of urea, you can solve for the molar mass of the unknown substance. Through cross-multiplication, as shown in the exercise solution, the molar mass of the unknown can be determined, making this an effective technique for molar mass determination when facing isotonic solutions.