A 279.6 mL sample of an aqueous solution at 25°C contains 91.6 mg of an unknown nonelectrolyte compound. If the solution has an osmotic pressure of 8.44 torr, what is the molar mass (in g/mol) of the unknown compound?

The molar mass of the compound is 720.8 g/mol

Explanation:

Let's apply the colligative property of Osmotic pressure to solve this.

Formula is π = M. R. T

where π is pressure (atm)

M is molarity (mol/L)

R, Universal Constant Gases

T, Absolute T° (T° in K = T° in C + 273)

Let's replace the dа ta:

8.44 Torr = M. 0.082 L. atm/mol. K. 298K

As we have the pressure in Torr, we must convert to atm, to work properly.

8.44 Torr. 1 atm / 760 Torr = 0.0111 atm

0.0111 atm = M. 0.082 L. atm/mol. K. 298K

0.0111 atm / (0.082 L. atm/mol. K. 298K) = M → 4.54*10⁻⁴ mol/L

So molarity is the moles of solute (mass (g) / molar mass) / volume (L)

Let's convert the volume to L → 279.6 mL. 1L / 1000 mL = 0.2796 L

4.54*10⁻⁴ mol/L. 02796 L = 1.27*10⁻⁴ moles

This moles are represented by the 91.6 mg, so let's convert the mass of solute from mg to g

91.6 mg. 1 g / 1000 mg = 0.0916 g

Molar mass → g/mol → 0.0916 g / 1.27*10⁻⁴ moles → 720.8 g/mol