Answer to Question #94698 in General Chemistry for julia lancaster

Question

A small cube of lithium (density = 0.535 g/cm3) measuring 1.0 mm on each edge is added to 0.520 L of water.The following reaction occurs:
2Li(s)+2H2O(l)→2LiOH(aq)+H2(g)

What is the freezing point of the resultant solution, assuming that the reaction goes to completion?

Answer 1

Since 1.0 mm = 0.10 cm, the volume of the cube is v = 0.10^3 = 1X10^-3 cm^3

From the density, 1.0X10^-3 cm^3 X 0.535 g/cm^3 = 5.35X10^-4 g Li

Since the molar mass of Li is 6.94 g/mol, that cube contains: 5.35X10^-4 g X (1 mol/6.94 g) = 7.7X10^-5 moles Li.

Because according to the equation, 2 mol Li forms 2 mol LiOH, you will produce 7.7X10^-5 moles LiOH. Since this ionizes, into Li+ and OH- in solution, you will have 7.7X10^-5 X 2 = 1.54X10^-4 moles of ions in the solution.

Since the density of H2O is 1 g/mL, the mass of the water in the solution will be 0.470 kg.

Since the definition of molality is moles of solute particles per kilogram of solvent, the molality of ions in the solution will be:

1.54X10^-4 mol / 0.470 kg = 3.28X10^-4 molal

The change in the freezing point of water is given by:

Delta T = - kf m and kf = 0.51 C/m, the change in the freezing point of this solution is:

Delta T = - (0.51 C/m) X 3.28X10^-4 m = -1.67X10^-4 C

It would be rather tricky to actually be able to measure this tiny change in the freezing point of the solution.