Question

20.0 L of an aqueous solution of silver nitrate (AgNO3). 10.0 L of a 0.0500 mol / L solution of calcium iodide (CaI2) is added thereto. At the end, all the Ag + ions precipitated and the molar concentration of the I ions is 0.0100 mol / L.
a) What was the initial molar concentration of AgNO3?
b) What mass of silver iodide has precipitated?
c) What is the melting point of the final water (suppose 1 liter of solution contains 1 kg of water)?

Expected answers :

a) initial molar concentration of silver nitrate = 0.0350 mol / L

b) mass of solid = 164 g

c) -0.0930 C

a) initial molar concentration of silver nitrate = 0.0350 mol / L b) mass of solid = 164 g c), T 0 0930 C f = - °

Answer 1

a) no of mol of caI2 present = 10*0.05 = 0.5 mol

   concentration of I- in the solution = 2*0.5/30 = 3.334*10^-2 M

   1 mol Ag+ = 1 mol I-.

initial molar concentration of silver nitrate = 0.0334 M

b) mass of AgI precipitated = n*mWT

                            = 0.0334*235

                            = 7.85 g

C) DTf = i*kf*m

DTf = T0-Tf = 0-x

DTf = i*Kf*m

i = vanthoff factor of Ca(NO3)2 = 2

Kf = frreezing point constant of water = 1.86 C/m

molality(m) = (n)*(1000/W)

   m = 0.5*(1/30) = 0.0167 m

(0-x) = 2*1.86*0.0167

x = freezing point of solution = -0.0621 C