Question
Define boiling point. Write the formula to determine molar mass of a solute using freezing point depression method.
✓
Answer
The boiling point is the temperature at which the vapour pressure of a liquid becomes equal to the atmospheric pressure.
Depression of freezing point and molar mass of the solute:
When W2 gram of solute with molar mass M2 is dissolved in W1 gram of solvent, the molality m is given by
$\Delta T_f \propto m$
$m=\frac{\text { Mass of solute in } kg }{\text { Molar mass of solue in } kg \times \text { Mass of solvent in } kg }$
$m=\frac{W_2}{M_2 \cdot W_1}$
∴ The molar mass of a solute using freezing depression method is givenby,
$\Delta T_f \propto m$
$\Delta T_f=k_f \cdot \frac{ W _2}{ M _2} \times \frac{1000}{ W _1}$
Where $K_f$ - Cryoscopic constant
$W_2$ - mass of solute
$W _1$ - mass of solvent
$M _2$ - molecular mass of non-volatile solute
Depression of freezing point and molar mass of the solute:
When W2 gram of solute with molar mass M2 is dissolved in W1 gram of solvent, the molality m is given by
$\Delta T_f \propto m$
$m=\frac{\text { Mass of solute in } kg }{\text { Molar mass of solue in } kg \times \text { Mass of solvent in } kg }$
$m=\frac{W_2}{M_2 \cdot W_1}$
∴ The molar mass of a solute using freezing depression method is givenby,
$\Delta T_f \propto m$
$\Delta T_f=k_f \cdot \frac{ W _2}{ M _2} \times \frac{1000}{ W _1}$
Where $K_f$ - Cryoscopic constant
$W_2$ - mass of solute
$W _1$ - mass of solvent
$M _2$ - molecular mass of non-volatile solute
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