Determine the molal concentration of a table sugar (C₁₂H₂₂O₁₁, 342.3 g/mol) solution in water that has a freezing point of -2.10$\degree$C.(K_f =1.86$\degree$C/m for water)

Question

Quizplus · Accepted Answer

The freezing point depression formula is $\Delta T_f = K_f \times m$, where $\Delta T_f$ is the change in freezing point, $K_f$ is the freezing point depression constant for the solvent, and $m$ is the molality of the solution. Given that the freezing point of the solution is -2.10°C and the freezing point of pure water is 0°C, the change in freezing point ($\Delta T_f$) is 2.10°C. With $K_f$ for water being 1.86°C/m, we can solve for $m$ (molality) as follows: $m = \frac{\Delta T_f}{K_f} = \frac{2.10}{1.86} = 1.13$ m.

Determine the Molal Concentration of a Table Sugar (C12H22O11, 342 °\degree°

Determine the Molal Concentration of a Table Sugar (C₁₂H₂₂O₁₁, 342 $\degree$