Question

Mathematical expressions for depression in freezing point​

Answer 1

я тебя не понимаю чуть чуть. математические выражения для депрессии в точке замерзания?

Answer 2

The depression in freezing point ($\Delta T_f$) of a solvent can be calculated using Raoult's law and the formula:

$\Delta T_f = K_f \times m$

Where:

• $\Delta T_f$ is the depression in freezing point (in degrees Celsius or Kelvin).
• $K_f$ is the cryoscopic constant, which is specific to the solvent and is given in units of °C kg/mol or K kg/mol.
• $m$ is the molality of the solution, which is the amount of solute in moles divided by the mass of the solvent in kilograms.

The relationship between molality ($m$) and mass of the solvent ($m_s$) and molar mass of the solvent ($M_s$) is given by:

$m = \frac{n_{\text{solute}}}{m_{\text{s}} \times M_{\text{s}}}$

Where:

• $n_{\text{solute}}$ is the amount of solute in moles.
• $m_{\text{s}}$ is the mass of the solvent in kilograms.
• $M_{\text{s}}$ is the molar mass of the solvent in kg/mol.

Substituting this expression for $m$ into the first equation, you get the depression in freezing point ($\Delta T_f$) in terms of the mass of the solvent ($m_{\text{s}}$), the molar mass of the solvent ($M_{\text{s}}$), and the cryoscopic constant ($K_f$):

$\Delta T_f = \frac{K_f \times n_{\text{solute}}}{m_{\text{s}} \times M_{\text{s}}}$

This formula allows you to calculate the depression in freezing point when you know the cryoscopic constant, the amount of solute, the mass of the solvent, and the molar mass of the solvent.