What is the boiling point of an aqueous solution that has a vapor pressure of 21.0 torr at 25 ∘C? (P∘H2O=23.78 torr; Kb= 0.512 ∘
1 answer:
Answer:
103.76° C
Explanation:
Let's use Raoult's Law to determine the ration of the mole fraction of the solvent by substituting the given pressure values:
We then determine the molality of the solution:
#Compute the boiling point elevation as:
#Add to boiling point of solution:
Hence the boiling point of the aqueous solution is 103.76° C
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where S=67 m is the distance covered, vf=0 is the final velocity of the car, and vi=15 m/s is the initial velocity. From this we can find a:
2) Then, we can assume this is the value of acceleration that the car is able to reach. In fact, the force the brakes are able to apply is
This force will be constant, and since m is always the same, then a is the same even in the second situation.
3) Therefore, in the second situation we have a=-1.68 m/s^2. However, the initial velocity is different: vi=45 m/s. Using the same formula of point 1), we can calculate the distance covered by the car before stopping: