Calculate the vapor pressure lowering of water when 5.67 g of glucose, C6H12O6, is dissolved in 25.2 g of water at 25 ºC. The vapor pressure of water at 25º C is 23.8 mmHg. What is the vapor pressure of the solution?
mols glucose = g/molar mass
mols H2O = grams/molar mass
XH2O = n H2O/total mols, then
Psoln = XH2O*Ponormal
For lowering, use
delta Psolvent =Xsolute*Ponormal
100 gram solution of 10 gram of magnesium.klorida and 90 gram of water what is the vapor pressure lowering of a solution if p h2o=31 mmhg
To calculate the vapor pressure lowering of water when a solute is dissolved in it, we can use Raoult's Law. According to Raoult's Law, 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.
First, let's calculate the moles of glucose (C6H12O6) and water in the solution.
The molar mass of glucose (C6H12O6) is:
(6 carbon atoms x 12.01 g/mol) + (12 hydrogen atoms x 1.01 g/mol) + (6 oxygen atoms x 16.00 g/mol) = 180.18 g/mol
The number of moles of glucose is:
5.67 g / 180.18 g/mol = 0.0315 mol
The molar mass of water (H2O) is:
(2 hydrogen atoms x 1.01 g/mol) + (1 oxygen atom x 16.00 g/mol) = 18.02 g/mol
The number of moles of water is:
25.2 g / 18.02 g/mol = 1.399 mol
Next, let's calculate the mole fraction of water in the solution:
Mole fraction of water (Xwater) = moles of water / total moles of solute and solvent
Xwater = 1.399 mol / (0.0315 mol + 1.399 mol) = 0.9788
Now, we can use Raoult's Law to calculate the vapor pressure of the solution:
Vapor pressure of solution = Vapor pressure of pure solvent x Mole fraction of solvent
Vapor pressure of solution = 23.8 mmHg x 0.9788 = 23.278 mmHg
Therefore, the vapor pressure of the solution is approximately 23.278 mmHg.
To calculate the vapor pressure of the solution, we first need to determine the number of moles of glucose and water. Then we can use Raoult's law to find the vapor pressure lowering.
Step 1: Find the moles of solute (glucose):
Molar mass of glucose (C6H12O6) = (6 * 12.01) + (12 * 1.01) + (6 * 16.00) = 180.18 g/mol
Number of moles of glucose = mass / molar mass
Number of moles of glucose = 5.67 g / 180.18 g/mol
Step 2: Find the moles of solvent (water):
Molar mass of water (H2O) = (2 * 1.01) + 16.00 = 18.02 g/mol
Number of moles of water = mass / molar mass
Number of moles of water = 25.2 g / 18.02 g/mol
Step 3: Calculate the vapor pressure lowering using Raoult's law:
The vapor pressure lowering (∆P) is given by the equation:
∆P = Xsolute * Psolvent
where Xsolute is the mole fraction of the solute and Psolvent is the vapor pressure of the solvent.
Mole fraction of glucose (Xsolute) = moles of glucose / total moles
Mole fraction of glucose = (moles of glucose) / (moles of glucose + moles of water)
Vapor pressure of the solution = Vapor pressure of the solvent - ∆P
Let's plug in the values and calculate:
Number of moles of glucose = 5.67 g / 180.18 g/mol ≈ 0.0315 mol
Number of moles of water = 25.2 g / 18.02 g/mol ≈ 1.401 mol
Mole fraction of glucose = 0.0315 mol / (0.0315 mol + 1.401 mol) ≈ 0.0223
Vapor pressure of the solution = 23.8 mmHg - (0.0223 * 23.8 mmHg)
Now, let's calculate the value:
Vapor pressure of the solution ≈ 23.8 mmHg - (0.0223 * 23.8 mmHg)
Vapor pressure of the solution ≈ 23.3 mmHg (rounded to the nearest tenth)
Therefore, the vapor pressure of the solution is approximately 23.3 mmHg.