a mixture of toluene and water boils at 90 degrees. calculate the vapour pressure at this temperature. Assume the atmospheric pressure to be 760 mmHg
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To calculate the vapor pressure of a mixture of toluene and water at a given temperature, we need to use Raoult's law.
Raoult's law states that the vapor pressure of an ideal liquid solution is directly proportional to the mole fraction of each component in the solution.
The mole fraction (X) of a component is the ratio of the number of moles of that component to the total moles of all components in the mixture.
In this case, we need to find the mole fraction of both toluene and water in the mixture. Let's assume that the mole fraction of toluene is X_toluene and the mole fraction of water is X_water.
Since we're given that the mixture boils at 90 degrees Celsius, this implies that the vapor pressure of the mixture is equal to the atmospheric pressure (760 mmHg or 1 atm) at that temperature.
Therefore, we can use these values to set up the following equation:
P_toluene * X_toluene + P_water * X_water = P_total,
where P_toluene and P_water are the vapor pressures of toluene and water at 90 degrees Celsius, and P_total is the total vapor pressure of the mixture (which is equal to atmospheric pressure).
We can rearrange this equation to solve for P_total:
P_total = P_toluene * X_toluene + P_water * X_water.
To find the vapor pressures of toluene and water at 90 degrees Celsius, we can refer to vapor pressure tables or use the Antoine equation, which relates vapor pressure to temperature.
Once we have the vapor pressures and mole fractions, we can substitute the values into the equation to calculate the vapor pressure of the mixture at 90 degrees Celsius.