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.

Fine out you self hx

Ans. : frostbite is when exposure to cold temperatures causes freezing to the skin or other tissues. it most commonly affects the hands, feet, and face. the longer areas are exposed to cold, typically the worse the frostbite. the water temperature used for rewarming a frostbitten part should be : (102 -105) f degrees.