Question

In a laboratory experiment, students synthesized a new compound and found that when 11.01 grams of the compound were dissolved to make 158.1 mL of a ethanol solution, the osmotic pressure generated was 1.91 atm at 298 K. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound?

Answer 1

892 g/mol

Step-by-step explanation:

Step 1: Given and required data

• Mass of the nonvolatile and a non-electrolyte solute: 11.01 g

• Volume of solution: 158.1 mL = 0.1581 L

• Osmotic pressure (π): 1.91 atm

• Ideal gas constant (R): 0.0821 atm.L/mol.K

• Absolute temperature (T): 298 K

Step 2: Calculate the molarity (M) of the solution

The osmotic pressure is a colligative property that can be calculated using the following expression.

π = M × R × T

M = π / R × T

M = 1.91 atm / (0.0821 atm.L/mol.K) × 298 K

M = 0.0781 M

Step 3: Calculate the molecular weight of the solute

We will use the following expression for molarity.

M = mass of solute / molecular weight of solute × liters of solution

molecular weight of solute = mass of solute / M × liters of solution

molecular weight of solute = 11.01 g / (0.0781 mol/L) × 0.1581 L

molecular weight of solute = 892 g/mol