1.00 of an aqueous solution contains 1.52g of a compound used in antifreeze. If the osmotic pressure of this solution at 20.0 degree C is 448 torr, calculate the molar mass of the antifreeze compound
1.00 what? grams? molar? L?
To calculate the molar mass of the antifreeze compound, we can use the equation:
Molar mass = (Molarity × R × Temperature) / Osmotic pressure
- Molarity (M) is the concentration of the solution, given as 1.00
- R is the ideal gas constant, approximately 0.0821 L·atm/(K·mol)
- Temperature (T) is given as 20.0 degrees Celsius, which needs to be converted to Kelvin (K). The conversion formula is: Kelvin = Celsius + 273.15
- Osmotic pressure (π) is given as 448 torr
Let's plug in the values:
Molar mass = (1.00 M × 0.0821 L·atm/(K·mol) × (20.0 + 273.15 K)) / 448 torr
First, let's convert torr to atm:
1 atm = 760 torr
Osmotic pressure in atm = 448 torr / 760 torr/atm
Now, substitute the values and solve for the molar mass:
Molar mass = (1.00 M × 0.0821 L·atm/(K·mol) × (20.0 + 273.15 K)) / (448 torr / 760 torr/atm)
Molar mass = (1.00 M × 0.0821 L·atm/(K·mol) × 293.15 K) / (448 torr / 760 torr/atm)
Molar mass = (1.00 M × 0.0821 L·atm/(K·mol) × 293.15 K) / (448/760)
Molar mass = (1.00 M × 0.0821 L·atm/(K·mol) × 293.15 K) / (0.59)
Finally, multiply the result by the molarity (concentration) of the solution, which in this case is 1.52g of the compound, to get the molar mass in grams per mole (g/mol).