Assuming that Raoults law is followed what would be the vapour pressure of a solution formed when 40g of sugar (molecular mass 342g) is dissolved in 360ml of water at 25oC? At 250C,the vapour pressure of water is 20mm and the density is 1g/ml.
To find the vapor pressure of the solution using Raoult's law, we need to know the mole fraction of the solute and the vapor pressure of the pure solvent.
Step 1: Calculate the number of moles of sugar:
Molar mass of sugar = 342g/mol
Mass of sugar = 40g
Number of moles of sugar = Mass of sugar / Molar mass of sugar
Number of moles of sugar = 40g / 342g/mol
Step 2: Calculate the number of moles of water:
Density of water = 1g/ml
Volume of water = 360ml
Mass of water = Density of water x Volume of water
Mass of water = 1g/ml x 360ml
Mass of water = 360g
Number of moles of water = Mass of water / Molar mass of water
Number of moles of water = 360g / 18g/mol
Step 3: Calculate the mole fraction of the solute:
Mole fraction of solute = Moles of sugar / (Moles of sugar + Moles of water)
Step 4: Calculate the vapor pressure using Raoult's law:
Vapor pressure of the solution = Mole fraction of solute x Vapor pressure of pure solvent
Given that the vapor pressure of water is 20mm Hg, we can now calculate the vapor pressure at 25°C:
Step 5: Convert temperature from °C to K:
Temperature in K = 25°C + 273.15 = 298.15K
Now we can perform the calculations:
Number of moles of sugar = 40g / 342g/mol ≈ 0.117 mol
Number of moles of water = 360g / 18g/mol = 20 mol
Mole fraction of solute = 0.117 mol / (0.117 mol + 20 mol) ≈ 0.0058
Vapor pressure of the solution = 0.0058 x 20mm Hg ≈ 0.116mm Hg
Therefore, at 25°C, the vapor pressure of the solution would be approximately 0.116mm Hg.
To determine the vapor pressure of a solution using Raoult's law, we need to consider the mole fraction of the solute and the vapor pressure of the solvent.
1. First, let's calculate the number of moles of sugar (solvent) dissolved in 360 mL of water.
Number of moles of sugar = mass of sugar / molar mass of sugar
= 40 g / 342 g/mol
≈ 0.117 mol
2. Next, calculate the mole fraction of sugar in the solution.
Mole fraction of sugar = moles of sugar / total moles of solute and solvent
= 0.117 mol / (0.117 mol + 0.36 mol) [since water has a density of 1g/ml and 360 ml is equal to 360 g]
≈ 0.245
3. The vapor pressure of pure water at 25oC is not provided, so we cannot use Raoult's law directly at this temperature. Instead, we can use the given information at 250C to estimate the vapor pressure at 25oC using the Clausius-Clapeyron equation:
ln(P2/P1) = (ΔHvap/R) * [(1/T1) - (1/T2)]
Where:
P1 = vapor pressure at 25oC (unknown)
P2 = vapor pressure at 250C (20 mm)
T1 = temperature in Kelvin at 25oC (25 + 273.15)
T2 = temperature in Kelvin at 250C (250 + 273.15)
ΔHvap = enthalpy of vaporization of water (~40.7 kJ/mol)
R = ideal gas constant (8.314 J/(mol·K))
Rearranging the equation to solve for P1:
P1 = P2 * exp[(ΔHvap/R) * [(1/T1) - (1/T2)]]
P1 = 20 mmHg * exp[(40700 J/mol / (8.314 J/(mol·K))) * [(1/(25 + 273.15) K) - (1/(250 + 273.15) K)]]
≈ 3.17 mmHg (approximately)
4. Finally, we can use Raoult's law to determine the vapor pressure of the solution:
Vapor pressure of the solution = Vapor pressure of water * mole fraction of water
= 3.17 mmHg * (1 - 0.245)
≈ 2.39 mmHg
Therefore, assuming Raoult's law is followed, the vapor pressure of the solution formed when 40g of sugar is dissolved in 360 mL of water at 25oC would be approximately 2.39 mmHg.
mols sugar = grams/molar mass = ?
mols H2O = grams/molar mass = ?
XH2O = mols H2O/total mols = ?
pH2O = XH2O*PoH2O
Post your work if you get stuck.