What weight of solute (M,wt =60) is required to be dissolved in 180gm of water to reduce vapour pressure to 4/5th of pure water?
Does it make a difference what the vapor pressure is? Here is what you use.
Psoln = Xsoln*Po/sup>solvent
To find the weight of solute (Molecular weight = 60) required to be dissolved in 180g of water to reduce the vapor pressure to 4/5th of pure water, we need to use Raoult's law.
Raoult's law states that the partial vapor pressure of a component in an ideal solution is directly proportional to its mole fraction in the solution. Mathematically, it can be represented as:
P_A = X_A * P°_A
where P_A is the partial vapor pressure of component A, X_A is the mole fraction of component A, and P°_A is the vapor pressure of pure component A.
In this case, we want to reduce the vapor pressure of the water to 4/5th of pure water. So, the partial vapor pressure of water in the solution will be:
P_A = (4/5) * P°_water
Now, we need to calculate the mole fraction of water in the solution. The mole fraction is defined as the ratio of the number of moles of a component to the total moles in the system.
Moles of water = Mass of water / Molecular weight of water
= 180g / 18g/mol (molecular weight of water)
= 10 moles
Total moles in the system = moles of water
Hence, the mole fraction of water in the solution is 1.
Using Raoult's law, we can set up the following equation:
(4/5) * P°_water = X_water * P°_water
Since X_water = 1, the equation becomes:
(4/5) * P°_water = 1 * P°_water
Canceling out P°_water from both sides, we get:
4/5 = 1
This is not possible and implies that we cannot achieve the desired vapor pressure reduction with the given conditions. The weight of solute needed cannot be calculated.
It's important to note that Raoult's law assumes ideal solutions, and deviations can occur in real-world scenarios. In cases where Raoult's law doesn't apply, one must consider other factors like interactions between solute and solvent molecules or the presence of non-volatile solutes.