What is the freezing point of a 0.975m aqueous solution of calcium nitrate (Ca(NO3)2)? Assume 100% disassociation.
Calcium nitrate (Ca(NO3)2) is a salt that completely dissociates in water into calcium ions (Ca2+) and nitrate ions (NO3-).
Since the solution is 0.975m, it means there is 0.975 moles of Ca(NO3)2 per liter of solution. This also means there is 0.975 moles of calcium ions and 1.95 moles of nitrate ions per liter of solution.
The Van't Hoff factor for calcium nitrate is 3, as it dissociates into 1 calcium ion and 2 nitrate ions.
Now, we can calculate the freezing point depression using the formula:
ΔTf = i * Kf * m,
where ΔTf is the freezing point depression, i is the Van't Hoff factor, Kf is the cryoscopic constant for water (1.86°C/m), and m is the molality of the solution.
Plugging in the values, we get:
ΔTf = 3 * 1.86 * 0.975 = 5.58°C
Therefore, the freezing point of a 0.975m aqueous solution of calcium nitrate is decreased by 5.58°C.
Since the normal freezing point of water is 0°C, the freezing point of the solution would be -5.58°C.