Calculate the approximate boiling point of a mixture of bromobenzene and water
at atmospheric pressure. A table of vapor pressure of water and bromobenzene at
various temperatures is given.
Vapor Pressures (mmHg)
Temperature(c) Water Bromobenzene
93 588 110
94 611 114
95 634 118
96 657 122
97 682 127
98 707 131
99 733 136
Solutions boil when their vapor pressure = atmospheric pressure, which I assume in this case 760 mm.
At 95 total P = 752
at 96 total P = 779 so looks like 95+ (or 96-). You can interpolate between those two values if you want to get closer but the problem says to estimate so I would answer "a little less than 96".
Well, calculating the boiling point of a mixture is quite complex, but don't worry, I'm here to clown around and help you out!
Let's start by comparing the vapor pressures of water and bromobenzene at different temperatures. As the temperature increases, so does the vapor pressure for both substances. This indicates that both water and bromobenzene will start to boil at higher temperatures.
Now, since we're looking for the boiling point of the mixture, we need to find the temperature at which the total vapor pressure of the mixture is equal to the atmospheric pressure.
But since I'm a funny bot, I'll tell you a secret: the boiling point dance-off between water and bromobenzene can get pretty heated! So, let's turn up the heat and get to the approximate boiling point.
Looking at the data, we can see that at 99°C, the vapor pressure of water is 733 mmHg, and the vapor pressure of bromobenzene is 136 mmHg.
To determine the approximate boiling point, we need to find the temperature at which the sum of the two vapor pressures equals the atmospheric pressure (which is typically around 760 mmHg at sea level). So, we'll have to use our math skills and perform some calculations.
If we add the vapor pressures of water and bromobenzene at 99°C, we get 733 + 136 = 869 mmHg.
Since our atmospheric pressure is around 760 mmHg, we know that the boiling point of the mixture must be higher than 99°C for the vapor pressure to be above atmospheric pressure.
Therefore, let's shake things up a bit and conclude that the approximate boiling point of the mixture of bromobenzene and water at atmospheric pressure is higher than 99°C.
Remember, this is just an approximation, so the real boiling point may be slightly different. But hey, at least we had some fun along the way, right?
To calculate the approximate boiling point of a mixture of bromobenzene and water at atmospheric pressure, we can use the fact that the boiling point is the temperature at which the vapor pressure of the liquid is equal to the external pressure (atmospheric pressure in this case).
From the given table, we can see that the vapor pressures of water and bromobenzene increase with increasing temperature.
To determine the boiling point, we need to find the temperature at which the total vapor pressure of the mixture is equal to the atmospheric pressure, which is typically around 760 mmHg.
Let's calculate the total vapor pressure at each temperature:
Temperature(c) Total Vapor Pressure (mmHg)
93 588 + 110 = 698
94 611 + 114 = 725
95 634 + 118 = 752
96 657 + 122 = 779
97 682 + 127 = 809
98 707 + 131 = 838
99 733 + 136 = 869
From the table, it can be observed that the total vapor pressure of the mixture exceeds atmospheric pressure at 96°C. Therefore, the approximate boiling point of the mixture of bromobenzene and water at atmospheric pressure is approximately 96°C.
To calculate the approximate boiling point of a mixture of bromobenzene and water at atmospheric pressure, we need to find the temperature where the total vapor pressure equals the atmospheric pressure.
The atmospheric pressure at sea level is approximately 760 mmHg.
First, let's find the approximate vapor pressure of water and bromobenzene at the boiling point by interpolating between the given data points.
For water:
At 97ºC, the vapor pressure is 682 mmHg.
At 98ºC, the vapor pressure is 707 mmHg.
To interpolate the vapor pressure at the boiling point of water:
Vapor pressure at 97ºC + ((Vapor pressure at 98ºC - Vapor pressure at 97ºC) * (Boiling point - 97ºC) / (98ºC - 97ºC))
For bromobenzene:
At 131ºC, the vapor pressure is 707 mmHg.
At 136ºC, the vapor pressure is 733 mmHg.
To interpolate the vapor pressure at the boiling point of bromobenzene:
Vapor pressure at 131ºC + ((Vapor pressure at 136ºC - Vapor pressure at 131ºC) * (Boiling point - 131ºC) / (136ºC - 131ºC))
To calculate the boiling point where the total vapor pressure equals atmospheric pressure:
Boiling point = 97ºC + ((131ºC - 97ºC) * ((760 mmHg - Vapor pressure of water) / (Vapor pressure of bromobenzene - Vapor pressure of water)))
Now, let's calculate the vapor pressure of water and bromobenzene at the boiling point:
Vapor pressure of water at the boiling point:
Vapor pressure at 97ºC + ((Vapor pressure at 98ºC - Vapor pressure at 97ºC) * (Boiling point - 97ºC) / (98ºC - 97ºC))
= 682 mmHg + ((707 mmHg - 682 mmHg) * (Boiling point - 97ºC) / (98ºC - 97ºC))
Vapor pressure of bromobenzene at the boiling point:
Vapor pressure at 131ºC + ((Vapor pressure at 136ºC - Vapor pressure at 131ºC) * (Boiling point - 131ºC) / (136ºC - 131ºC))
= 707 mmHg + ((733 mmHg - 707 mmHg) * (Boiling point - 131ºC) / (136ºC - 131ºC))
Finally, let's calculate the boiling point:
Boiling point = 97ºC + ((131ºC - 97ºC) * ((760 mmHg - Vapor pressure of water) / (Vapor pressure of bromobenzene - Vapor pressure of water)))