water has a vapour pressure of 23.76 torr at 25C.what is the vapour pressure of a solution of sucrose ,if its mole fraction is 0.25?
I assume you mean Xsucrose = 0.25 which makes XH2O = 0.75.
Then Psoln = XH2O*Ponormal
Psoln (in torr) = 0.75*23.76
To determine the vapour pressure of a solution of sucrose, you will need to use Raoult's law, which states that the vapour pressure of a solution is directly proportional to the mole fraction of the solvent present in the solution.
In this case, the solvent is water, and the mole fraction of sucrose (solute) is given as 0.25.
Raoult's law can be written as:
P = P° * X
Where P is the vapour pressure of the solution, P° is the vapour pressure of the pure solvent (water), and X is the mole fraction of the solvent.
Now, we have the vapour pressure of water (P°) at 25°C, which is 23.76 torr. The mole fraction of sucrose (X) is given as 0.25.
Plugging in the values into Raoult's law, we get:
P = 23.76 torr * 0.25
P = 5.94 torr
Therefore, the vapour pressure of the solution of sucrose with a mole fraction of 0.25 is 5.94 torr.