Question

Which solution has the same boiling point as 0.25 mol CaCl2 dissolved in 1000 g water?

Answer 1

The boiling point of water at 1 atm is 100 degrees celsius. However, when water is added with another substance the boiling point of it rises than when it is still a pure solvent. This called boiling point elevation, a colligative property. The equation for the boiling point elevation is expressed as the product of the ebullioscopic constant (0.52 degrees celsius / m) for water), the vant hoff factor and the concentration of solute (in terms of molality). 

 ΔT(CaCl2) = i x K x m = 3 x 0.52 x 0.25 = 0.39 °C
 ΔT(Sucrose) = 1 x 0.52 x 0.75 = 0.39 °C
 ΔT(Ethylene glycol) = 1 x 0.52 x 1 = 0.52 °C
 ΔT(CaCl2) = 3 x 0.52 x 0.50 = 0.78 °C
 ΔT(NaCl) = 2 x 0.52 x 0.25 = 0.26 °C

Thus, from the calculated values, we see that 0.75 mol sucrose dissolved on 1 kg water has the same boiling point with 0.25 mol CaCl2 dissolved in 1 kg water.