(Solved): 40.00 grams of K3PO4 (potassium phosphate, molar mass =212g/mol ) are dissolved in 500.00 ...

$$40.00$$ grams of $${\mathrm{K}}_3{\mathrm{PO}}_4$$ (potassium phosphate, molar mass $$=212\mathrm{g}/\mathrm{mol}$$ ) are dissolved in $$500.00$$ grams of $${\mathrm{H}}_2\mathrm{O}$$ (water, molar mass $$=18\mathrm{g}/\mathrm{mol}$$ ). The volume of the homogeneous mixture is $$535.0\mathrm{mL}$$. Match the colligative property with the corresponding numeric value. The dissolution equation for $${\mathrm{K}}_3{\mathrm{PO}}_4$$ in $${\mathrm{H}}_2\mathrm{O}$$ is: $${\mathrm{K}}_3{\mathrm{PO}}_{\mathrm{s}}(\mathrm{s})\underset{{\mathrm{H}}_2\mathrm{O}}{?}3{\mathrm{K}}^{+}(\mathrm{aq})+{\mathrm{PO}}_4^{3?(\mathrm{aq})}$$ For every unit of $${\mathrm{K}}_3{\mathrm{PO}}_4$$ that dissolves, 4 particles are formed. The formulas are moles $${\mathrm{K}}_3{\mathrm{DO}}_4=\frac{\text{ mass }{\mathrm{K}}_3{\mathrm{PO}}_4}{\mathrm{MMK}{\mathrm{K}}_3{\mathrm{PO}}_4}$$ moles $${\mathrm{H}}_2\mathrm{O}=\frac{\text{ mass }{\mathrm{H}}_2\mathrm{O}}{\mathrm{MM}{\mathrm{H}}_2\mathrm{O}}$$ $$\underset{\begin{array}{c}\text{ Osmatic }\\ \text{ pressure }\end{array}}{}$$ of solution $$=\frac{\left(\frac{\text{ mass }}{MM}\right)K_{3}P{0}_4}{L\text{ solution }}\times \left(\frac{\text{ (tparticles }}{\text{ compound }}\right)\times 0.0821\frac{L}{\text{ molatim }}\times \underset{\begin{array}{c}\text{ in }\\ \text{ kelvin }\end{array}}{T}$$ 1. boiling point of solution, in $${}^{?}\mathrm{C}$$ 2. freezing point of solution, in $${}^{?}\mathrm{C}$$ 3. vapor pressure of solution at $$22.0$$ $${}^{?}\mathrm{C}$$, in $$\mathrm{mm}\mathrm{Hg}$$ vapor pressure of pure $${\mathrm{H}}_2\mathrm{O}=19.8$$ $$\mathrm{mm}\mathrm{Hg}$$ 4. osmotic pressure of solution at $$22.0$$ $${}^{?}\mathrm{C}$$, in atm