Question

When 19.1 g of a nonvolatile and nonelectrolyte solute of unknown identity are dissolved in 150 ml of water at room temperature of 25 °C, a solution is obtained whose osmotic pressure is 23 atm. the density of the previous solution is 1.06 g/ml. Choose which of the following statements is correct:

1) The molar mass of the solute is 50 g/mol
2) The molar mass of the solute is 100 g/mol
3) The molar mass of the solute is 150 g/mol
4) The molar mass of the solute cannot be determined

Answer 1

By applying the osmotic pressure formula and solving for the molarity of the solution, we can calculate the molar mass of the solute to be approximately 135 g/mol, which indicates that the correct statement is (3) The molar mass of the solute is 150 g/mol.

Step-by-step explanation:

To determine the correct statement regarding the molar mass of the solute, we first need to use the osmotic pressure formula: II = MRT, where II is the osmotic pressure, M is the molarity of the solution, R is the ideal gas constant (0.0821 L·atm/K·mol), and T is the temperature in Kelvin. With the given osmotic pressure (23 atm), temperature (25 °C or 298 K), and volume of solution (150 mL or 0.150 L), we can calculate the molarity (M) of the solution. Then, using the formula mass (grams of solute) divided by molar mass (grams per mole), we can solve for molar mass.

First, calculate the molarity: M = II / (RT) = 23 atm / (0.0821 L·atm/K·mol * 298 K) = 0.943 mol/L. Now, calculate the molar mass: molar mass = 19.1 g / (0.943 mol/L * 0.150 L) = 135 g/mol. Therefore, the correct statement is (3) The molar mass of the solute is 150 g/mol, as it is the closest to our calculated value.