Is the boiling point of 0.01 m potassium fluoride solution higher or lower than that of 0.01 m glucose solution? Explain
delta T = i*Kb*m
i = van't Hoff factor
i = 1 for glucose
i = 2 for KF
Take a look at the formula. For the same m and the same Kb, the only difference is i.
So i*Kb*m must be a greater number for which one. And the greater number for delta T + 100 (the normal boiling point for H2O) means a higher boiling point.
Because KF has two ion
The boiling point of a solution depends on the total concentration of solute particles present in the solution. This is determined by the number of solute particles and their nature. In general, the more solute particles present in a solution, the higher its boiling point.
In a 0.01 m potassium fluoride (KF) solution, potassium fluoride dissociates into K+ cations and F- anions. Therefore, for each molecule of KF, it generates two solute particles. Thus, the effective concentration of solute particles in the KF solution is 0.02 M.
On the other hand, in a 0.01 m glucose (C6H12O6) solution, glucose remains as an intact molecule and does not dissociate or ionize. Therefore, for each molecule of glucose, it remains as a single solute particle. Thus, the effective concentration of solute particles in the glucose solution is 0.01 M.
Comparing the two solutions, the 0.01 m potassium fluoride solution has a higher effective concentration of solute particles (0.02 M) compared to the 0.01 m glucose solution (0.01 M). Since boiling point elevation is directly proportional to the concentration of solute particles, the boiling point of the potassium fluoride solution will be higher than that of the glucose solution.
In conclusion, the boiling point of a 0.01 m potassium fluoride solution is higher than that of a 0.01 m glucose solution due to the higher effective concentration of solute particles in the potassium fluoride solution.
To determine whether the boiling point of a 0.01 m potassium fluoride (KF) solution is higher or lower than that of a 0.01 m glucose (C6H12O6) solution, we need to consider the properties and behavior of each solute.
The boiling point of a solution is influenced by the number of solute particles present in the solution, which is directly proportional to the concentration of the solution. Therefore, to compare the boiling points, we need to examine the number of solute particles in each solution.
In a solution, potassium fluoride (KF) dissociates into K+ and F- ions. This means that each KF molecule breaks into two particles when dissolved. Therefore, a 0.01 molar solution of KF will contain twice the number of solute particles as the original molarity. So, the effective concentration of K+ and F- ions in the solution would be 0.02 M.
On the other hand, glucose (C6H12O6) does not dissociate or ionize when dissolved in water. It remains as individual glucose molecules in solution. Therefore, a 0.01 molar solution of glucose would maintain the same concentration of solute particles, that is, 0.01 M.
Since the effective concentration of solute particles in the KF solution (0.02 M) is higher than that in the glucose solution (0.01 M), the boiling point of the KF solution would be higher.
To summarize, the boiling point of the 0.01 m potassium fluoride solution would be higher than that of the 0.01 m glucose solution due to the higher concentration of solute particles resulting from the dissociation of KF into ions.