determine the vapor pressure lowering at 27 C of a solution of 750g of sucrose, dissolved in 180g of water. The vapor pressure of pure water at 27 C is 26.7 torr. Assume the solution is ideal
Hey it's me again, I forgot to type that the "Sucrose is nonvolatile,nionaizing solute in water" and the answer to it was 0.571 but what is the solution or how was that the answer? Please help me :)
I think you have made a typo. There is no way to get 0.571 (I assume that is torr) from the problem you posted. I think you have made a typo in the 750 g sucrose; I believe that must be 75.0 g sucrose I will assume my assumption is right.
mols sucrose = 75.0/342 = ?
mols H2O = 180/18 = ?
Xsucrose = mols sucrose/total mols.
Then delta P = Xsucrose*Po H2O and that comes out to about 0.572 torr.
To determine the vapor pressure lowering of a solution, we can use Raoult's Law, which states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution.
The mole fraction of the solvent (water in this case) can be calculated using the given masses of the solute (sucrose) and solvent (water).
Step 1: Calculate the moles of sucrose and water:
The molar mass of sucrose (C12H22O11) is approximately 342.30 g/mol.
Number of moles of sucrose = mass of sucrose / molar mass of sucrose
= 750 g / 342.30 g/mol
≈ 2.191 mol
The molar mass of water (H2O) is approximately 18.02 g/mol.
Number of moles of water = mass of water / molar mass of water
= 180 g / 18.02 g/mol
≈ 9.99 mol
Step 2: Calculate the mole fraction of water:
Mole fraction of water = moles of water / (moles of water + moles of sucrose)
= 9.99 mol / (9.99 mol + 2.191 mol)
≈ 0.82
Step 3: Use Raoult's Law to calculate the vapor pressure lowering:
The vapor pressure lowering (∆P) is given by the difference between the vapor pressure of pure water (P₀) and the vapor pressure of the solution (P).
∆P = P₀ - P
Given that the vapor pressure of pure water (P₀) at 27°C is 26.7 torr, we need to find the vapor pressure of the solution (P).
Using Raoult's Law:
P = P₀ * mole fraction of water
= 26.7 torr * 0.82
≈ 21.89 torr
∆P = 26.7 torr - 21.89 torr
≈ 4.81 torr
Therefore, the vapor pressure lowering at 27°C for the given solution is approximately 4.81 torr.