What is the density of an aqueous solution of potassium nitrate, that has a normal boiling point of 103.0 degrees Celsius, and an osmotic pressure of 122 at 25 degrees Celsius?

To determine the density of an aqueous solution of potassium nitrate, we can use the relationship between density, concentration, molar mass, and osmotic pressure.

First, let's convert the boiling point from Celsius to Kelvin. The boiling point in Kelvin is 103.0 + 273.15 = 376.15 K.

Next, we need to convert the osmotic pressure from atmospheres (atm) to Pascal (Pa). The conversion factor is 1 atm = 101325 Pa. So, the osmotic pressure is 122 * 101325 = 12,341,650 Pa.

The formula to calculate the density of a solution is:

Density (g/mL) = (Osmotic Pressure (Pa) * Molar Mass (g/mol) * 1,000) / (RT)

Where:
- Osmotic Pressure is the osmotic pressure in Pascal (Pa)
- Molar Mass is the molar mass of the solute in grams per mole (g/mol)
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (K)

The molar mass of potassium nitrate (KNO3) is:
[K (39.10 g/mol) + N (14.01 g/mol) + 3 O (16.00 g/mol)] = 101.10 g/mol

Now, we can substitute the values into the formula:

Density (g/mL) = (12,341,650 * 101.10 * 1,000) / (8.314 * 376.15)

Simplifying the expression gives:

Density = 3.95 g/mL

Therefore, the density of the aqueous solution of potassium nitrate is 3.95 grams per milliliter (g/mL).