The osmotic pressure of a solution of MgCl2 is 5.43 atm at 25 oC. How much of the salt is
dissolved in 250.0 ml of water?
To determine the amount of MgCl2 dissolved in 250.0 ml of water, we need to use the equation for osmotic pressure:
π = iMRT
Where:
π = osmotic pressure
i = van 't Hoff factor (number of particles per formula unit)
M = molarity of the solution
R = ideal gas constant
T = temperature in Kelvin
In this case, we have the osmotic pressure (π) as 5.43 atm at 25 oC. We need to convert the temperature to Kelvin by adding 273:
T = 25 + 273 = 298 K
The van 't Hoff factor (i) for MgCl2 is 3 because it dissociates into three particles in solution: one Mg²⁺ ion and two Cl⁻ ions.
The equation now becomes:
5.43 atm = (3)(M)(0.0821 L·atm/(mol·K))(298 K)
Now we can solve for M (molarity):
M = (5.43 atm) / [(3)(0.0821 L·atm/(mol·K))(298 K)]
M = 0.072 M
Molarity is defined as the number of moles of solute divided by the volume of solution in liters (mol/L). We have 250.0 ml of water, which is 0.250 L.
Now we can calculate the number of moles of MgCl2:
moles = (0.072 M)(0.250 L) = 0.018 mol
Therefore, 0.018 moles of MgCl2 are dissolved in 250.0 ml of water.