Question

-4 (3 pts.) If a 0.830 m aqueous solution freezes at -3.58 °C, what is the van't Hoff factor, i, of the solute? On Blackboard, please enter your response as a number with three significant figures le X.XX where the X's are numbers). Please do not use exponent notation or fractions. Also, please write your answer in the box.

Answer 1

Sol we know that atf = iktam where Alf - depression in freezing point si = freezing point of water - freezing point of solution. A Tf - oc - (-3.58°0) = 3.58°C Kf = depression in freeting Constant of water Kf = 1.86cm is vanthott's factor m = molarity of the solution - 0.830 m. ATf= iKfm 3.58 = x 1.86 x 0.830 m i= 3.58 1,86 X0-830 i = 2.318 So, the Van't Hoft factor = 2.32