Question

attempts left Check my work Enter your answer in the provided box. What is the freezing point of a solution that contains each of the following quantities of solute in 1.00 kg of water? 4.4 mol of Na3PO4

Answer 1

This is depression of freezing point and it is a colligative property of the solution.

Colligative properties solution that depends no. of solnte particle molecule in sointion, nature of chemical are properties on the ratio to the no. of and not on species. of g. solvent the Colligative properties - D Relative lower of vapour pressure 1) Elevation of boiling point 11) Depression of freezing point . W est Osmotic pressure for a given sointe - solvent ratio , all colligative properties are propotional to solnte molar mass. mass inversely

Depression freezing of point Formular: The freezing point of a pure solvent (let #9) is decreased by addition of a non-volatile solute ( let - Nach of 1 I = -i m where, AT = (solution) - [ (soment) It = freezing point = (ryoscopic constant izvan't Hoft factor of solvent ma molality.

is Na POA and solvent Now, is here solute water. of the first calculate the molality solution. Moles of solute = 4.4 mol Mass of solvent = 1kg

moles of solute Molality(m) = mass(in kg) of Solvent 4.4 mol 1kg = 4. 4 0 Kg = 4.4 m.

The formula for depression of freezing point is, ATư=i x Ke XM AT:= Change in freezing point i= Van't Hoff factor = 4 (three Nation and one PO43-) Ke=Freezing point constant = 1.86°Cm 1. m =molality of the solution=4.4 m AT= 4 x (1.86 °Cm-1) * 4.4m = 32.74°C Now normal freezing point of water is 0°C. So, freezing point of the solution is (0-32.74)°C = -32.74°C.