Answer:
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 data:
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
