To find the osmotic pressure (π) of the 2.36 M Cr(NO3)3 solution at 293 K, we can use the following formula:
π = MRT
Where M is the molarity of the solution, R is the gas constant (0.0821 L•atm/K•mol), and T is the temperature in Kelvin. Substituting the given values, we get:
π = (2.36 M) x (0.0821 L•atm/K•mol) x (293 K)
π = 58.12 atm
Therefore, the osmotic pressure of the 2.36 M Cr(NO3)3 solution at 293 K is 58.12 atm.
When 2 molecules of Cr(NO3)3 completely dissociate, they will form 3 ions in solution: one Cr3+ ion and three NO3- ions.
To find the molarity of the Cr(NO3)3 solution, we need to first calculate the number of moles of solute (Cr(NO3)3) dissolved in the solution:
Number of moles = mass / molar mass
The molar mass of Cr(NO3)3 is 241.99 g/mol (chromium has a molar mass of 51.996 g/mol, nitrogen has a molar mass of 14.007 g/mol, and oxygen has a molar mass of 15.999 g/mol; there are 3 nitrate ions, each with a molar mass of 62.004 g/mol). Substituting the given values, we get:
Number of moles = 300 g / 241.99 g/mol
Number of moles = 1.24 mol
Now that we know the number of moles of solute, we can use the following formula to find the molarity (M) of the solution:
M = moles of solute / volume of solution (in liters)
The volume of the solution is given in milliliters, so we need to convert it to liters by dividing by 1000:
M = 1.24 mol / (250 mL / 1000 mL/L)
M = 4.96 M
Therefore, the molarity of the Cr(NO3)3 solution is 4.96 M.