Question

A small cube of lithium (density = 0.535 g/cm3) measuring 1.0 mm on each edge is added to 0.550 L of water.The following reaction occurs: 2Li(s)+2H2O(l)→2LiOH(aq)+H2(g)

What is the freezing point of the resulting solution? Assume full dissociation of strong electrolytes and negligible solubility of hydrogen gas in water. The molal freezing-point-depression constant for water is 1.86°C/m.

I got -0.517 C but this is incorrect

Answer 1

volume = 1.0 mm3 = 0.001 cm3

density = 0.535 g/cm3

mass of Li = 0.001 x 0.535 = 5.35 x 10^-4

moles of Li = 5.35 x 10^-4 / 7 = 7.64 x 10^-5 mol

mass of water = 0.550 kg

molality = moles / mass of solvent

            = 7.64 x 10^-5 / 0.550

             = 1.39 x 10^-4 m

delta Tf = Kf x m

             = 1.86 x 1.39 x 10^-4

             = 2.6 x 10^-4

freezing point = - 0.00026 oC