The nonvolatile, nonelectrolyte saccharin, C7H5NO3S (183.2 g/mol), is soluble in ethanol, CH3CH2OH. How many grams of saccharin are needed to generate an osmotic pressure of 8.09 atm when dissolved in 246 mL of a ethanol solution at 298 K.

14.9 g

Explanation:

The osmotic pressure (π) is a colligative property that can be calculated through the following expression.

π = M * R * T

where

M: molarity

R: ideal gas constant

T: absolute temperature

We can use it to calculate the molarity.

M = π / R * T

M = 8.09 atm / (0.0821 atm. L/mol. K) * 298 K

M = 0.331 M

Given that the molarity is 0.331 M and the volume of the solution is about 246 mL (0.246 L), the moles of solute are:

0.246 L * 0.331 mol saccharin/L = 0.0814 mol saccharin

The molar mass of saccharin is 183.2 g/mol. The mass corresponding to 0.0814 moles is:

0.0814 mol * 183.2 g/mol = 14.9 g