The answer is 50.0 torr and I got 46.0 torr but idk what I did wrong.
Find the vapor pressure of a solution of 164 g of C3H8O3 and 338 mL of water at 39.8 degrees Celsius. Water's VP at 39.8 degrees celsius is 54.74 torr; it's density is 0.992 g/mL.
If you had shown what you did we could tell you what you did wrong.
moles C3H8O3 = grams/molar mass
moles H2O = (338 mL*0.992 g/mL)/molar mass
mole fraction H2O = moles H2O/total moles.
vap pressure water = moles fraction H2O x normal vap pressure water. I get 49.96 in a quickie calculation which rounds to 50.0 torr. You should do it more accurately than I.
To solve this problem, we can use Raoult's law, which states that at a given temperature, the vapor pressure of a solvent above a solution is directly proportional to the mole fraction of the solvent in the solution.
The mole fraction of a substance can be calculated by dividing the moles of the substance by the total moles in the solution.
First, let's calculate the moles of C3H8O3:
We are given the mass of C3H8O3, which is 164 g. To find the moles, we need to divide the mass by the molar mass of C3H8O3.
The molar mass is calculated by adding up the atomic masses of each element in the compound. In this case, we have:
C: 3 * 12.01 = 36.03 g/mol
H: 8 * 1.01 = 8.08 g/mol
O: 3 * 16.00 = 48.00 g/mol
Therefore, the molar mass of C3H8O3 is 36.03 + 8.08 + 48.00 = 92.11 g/mol.
Now, we can calculate the moles of C3H8O3:
moles = mass / molar mass = 164 g / 92.11 g/mol ≈ 1.78 mol
Next, let's calculate the moles of water:
We are given the volume of water, which is 338 mL. To find the moles, we need to divide the volume by the molar volume of water at the given temperature and pressure.
The molar volume of water at a specific temperature and pressure can be calculated using the ideal gas law equation:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.
First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15 = 39.8 + 273.15 = 313.95 K
Now, let's calculate the moles of water:
Using the ideal gas law, we can rearrange the equation to solve for n:
n = PV / (RT)
Given: P = 54.74 torr, V = 338 mL = 0.338 L, R = 0.0821 L·atm/(mol·K), and T = 313.95 K.
n = (54.74 torr * 0.338 L) / (0.0821 L·atm/(mol·K) * 313.95 K)
n ≈ 0.0095 mol
Next, let's calculate the mole fraction of the solvent (water):
Mole fraction of water = moles of water / total moles
Mole fraction of water = 0.0095 mol / (1.78 mol + 0.0095 mol) ≈ 0.0053
Now, we can use Raoult's law to find the vapor pressure of the solution:
vapor pressure = mole fraction of solvent * vapor pressure of pure solvent
vapor pressure = 0.0053 * 54.74 torr ≈ 0.290 torr
Therefore, the vapor pressure of the solution is approximately 0.290 torr.
It seems like you got 46.0 torr as the result, which is different from the correct answer of 0.290 torr. To identify the mistake, it would be helpful to review the steps and calculations you performed. Make sure you used the correct units and formulas, and double-check your calculations.