Assuming a 100% dissociation, what is the freezing point and boiling point of 2.59 m k3po4(aq).

Assuming 100% dissociation of K3PO4 you get each molecule of K3 PO4 dissociates into four particles: 3 K4 cations and 1 PO4 - anion. That means that Van't Hoff factor, i, is 4. i = 4. So, the decrease in freezing point of the water solution is Delta Tf = i * Kf * m. Where Kf is the molal cryoscopic constant of water = 1.86 °C /m; and the molality m = 2.59 m => Delta Tf = 4 * 1.86 °C / m * 2.59 m = 19.3 °C. And the freezing point is the normal freezing points less 19.3°C = 0°C - 19.3°C = - 19.3°C. The increase in the boiling point Delta Tb = i * kb * m = 4 * 0.512 °C /m * 2.59 m = 5.3°C => Tb = normal boiling point + 5.3°C = 100°C + 5.3°C = 105.3°C.

Ibelieve it is b. proton.hope my answer has you : )

2licl + 3o2 i yook the test and it was correct.