Answer to Question #94557 in Organic Chemistry for Karla

Question

A solution freezes at -2.30°C. At what temperature will it boil?

Answer 1

The сryoscopic constant for water is K_f =1.853 K·kg/mol.

The ebullioscopic constant for water is K_b = 0.512 K·kg/mol.

The molality of the solution b = ∆T_f / K_f = 2.30K / 1.853 K·kg/mol = 1.2412 mol/kg.

∆T_b = b × K_b = 1.2412mol/kg × 0.512 K·kg/mol = 0.6355 K ≈ 0.64K.

Therefore, it will boil at 100.64⁰C.