
The value of ${k_b}$ or ${K_f}$depends only on the type of solvent & not solute dissolved in it.
A. True
B. False
Answer
565.5k+ views
Hint: ${K_b}$ is the molal elevation in boiling point constant whereas ${K_f}$ is the molal depression in freezing point constant. These constants are different for different solvents.
Complete answer
It is the given statement that the value of ${K_b}$ and ${K_f}$ depends only on the type of the solvent and not the solute dissolved in it.
This is because both the molal elevation in boiling point constant i.e. ${K_b}$ and molal depression in freezing point constant i.e. ${K_f}$ are the characteristics of solvent and not the solute. Both ${K_b}$ and ${K_f}$ are independent of the solute.
${K_b}$ refers to the molal elevation constant. When one mole of solute is dissolved in 1000g (1Kg) of the solvent then the elevation produced in the boiling point is known as the molal elevation constant. The molal elevation constant is also known as ebullioscopic constant.
${K_f}$ refers to the molal depression constant. When one mole of solute is dissolved in 1000g (1Kg) of the solvent then the depression produced in the freezing point is known as the molal depression constant. Molal depression constant is also called cryoscopic constant. Molal depression constant is expressed as;
\[\;\Delta {T_f} = {\rm{ }}{K_f} \times {\rm{ }}m\];
where $\Delta {T_f}$= depression in freezing point,${K_f}$= molal depression constant,$m$= molality of solution
The molal elevation constant and the molal depression constant depend only upon the solvent of the solution. Both are the colligative properties of a solution. We know that molality is actually the number of moles of solute upon the weight of the solvent.
Hence, option (A) True, is the correct option.
Note:
Thus, the value of ${K_b}$and ${K_f}$ depends only on the type and nature of solvent. Both the molal elevation constant and the molal depression are totally independent of the nature of the solute. Also, from the expressions for ${K_b}$ and ${K_f}$ it is clear that both of them depend upon the nature of the solvent.
Complete answer
It is the given statement that the value of ${K_b}$ and ${K_f}$ depends only on the type of the solvent and not the solute dissolved in it.
This is because both the molal elevation in boiling point constant i.e. ${K_b}$ and molal depression in freezing point constant i.e. ${K_f}$ are the characteristics of solvent and not the solute. Both ${K_b}$ and ${K_f}$ are independent of the solute.
${K_b}$ refers to the molal elevation constant. When one mole of solute is dissolved in 1000g (1Kg) of the solvent then the elevation produced in the boiling point is known as the molal elevation constant. The molal elevation constant is also known as ebullioscopic constant.
${K_f}$ refers to the molal depression constant. When one mole of solute is dissolved in 1000g (1Kg) of the solvent then the depression produced in the freezing point is known as the molal depression constant. Molal depression constant is also called cryoscopic constant. Molal depression constant is expressed as;
\[\;\Delta {T_f} = {\rm{ }}{K_f} \times {\rm{ }}m\];
where $\Delta {T_f}$= depression in freezing point,${K_f}$= molal depression constant,$m$= molality of solution
The molal elevation constant and the molal depression constant depend only upon the solvent of the solution. Both are the colligative properties of a solution. We know that molality is actually the number of moles of solute upon the weight of the solvent.
Hence, option (A) True, is the correct option.
Note:
Thus, the value of ${K_b}$and ${K_f}$ depends only on the type and nature of solvent. Both the molal elevation constant and the molal depression are totally independent of the nature of the solute. Also, from the expressions for ${K_b}$ and ${K_f}$ it is clear that both of them depend upon the nature of the solvent.
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