Two solutions are prepared using the same solute: solution a.0.27 g of the solute dissolves in 27.4 g of t-butanol solution b.0.23 g of the solute dissolves in 24.8 g of cyclohexane which solution has the greatest freezing point change?

The freezing point depression is calculated through the equation,

     ΔTf = (Kf)(m)

where ΔTf is freezing point depression, Kf is the molal freezing point depression and m is the molality of the solute. Using the givens above and the equation,

a. Let x be the molar mass of the solute. The Kf of t-butanol is 9.1 °C/m.
      ΔTf = (9.1)(0.27/x)(1/0.0274)
      ΔTf = 89.67/x

b. Let x be the molar mass of the same solute. Kf value of 20.1 °C/m
     ΔTf = (20.1)(0.23/x)(1/0.0248)
     ΔTf = 186.4/x

The answer would be letter B. 

First balance the equation: you get 3Fe + 4H2O = Fe3O4 +4H2O

Then set up the equations according to the spidey map below.

x 10^25 molec. H2O/ 1 x x x

Multiply the top all together, then divide by the whole bottom half.

154.4 g H2

136 g Fe /1 x x x

Multiply the top all together, then divide by the whole bottom half.

4.95 x 10^20 Formula units (fu) Fe3O4

8.25 x 10^23 molec. Fe/ 1 x x

Multiply the top all together, then divide by the whole bottom half.

1.83 mol H2