How many grams of a nonvolatile compound B (molar mass = 97.8 g/mol) would need to be added to 250 g of water to produce a solution with a vapor of 23.756 torr? The vapor pressure of water at this temperature is 42.362 torr. Please round the answer to a whole number.
Psoln = XH2O*PoH2O
23.756 = X*42.362
Solve for X = mole fraction H2O needed. That's approx 0.6
Then X = mole fraction B must be 1 - 0.6 = approx 0.4
mols H2O = g/molar mass = 250/18 = about 14 (approximately)
XB = mols B/(mols B + mols H2O)
0.4 = mols B/(mols B + 14). Solve for mols B.
Then mols B = grams B/molar mass B. You know mols B and molar mass B, solve for grams B. Remember the numbers I've posted are approximate only. Post your work if you get stuck.
To solve this problem, we will use Raoult's law, which states that the vapor pressure of a component in a solution is proportional to its mole fraction. The mole fraction of compound B can be calculated by dividing the number of moles of B by the total number of moles in the solution.
Let's break down the information given:
Molar mass of compound B (MB) = 97.8 g/mol
Mass of water (MW) = 250 g
Vapor pressure of water (P°water) = 42.362 torr
Vapor pressure of the solution (P°solution) = 23.756 torr
First, we need to calculate the mole fraction of water (Xwater). We can do this by dividing the moles of water by the total moles of the solution.
Step 1: Calculate the moles of water
Number of moles of water (nwater) = mass of water (MW) / molar mass of water (Molar mass of water = 18.015 g/mol)
nwater = 250 g / 18.015 g/mol
Step 2: Calculate the total moles in the solution (ntotal)
ntotal = nwater + nB
Let's calculate the number of moles of water and the total moles in the solution:
nwater = 250 g / 18.015 g/mol = 13.87 mol
ntotal = 13.87 mol
Now, we need to calculate the mole fraction of water (Xwater):
Xwater = nwater / ntotal
Xwater = 13.87 mol / 13.87 mol = 1
Since compound B is nonvolatile, its vapor pressure in the solution will be zero. Therefore, the mole fraction of compound B (XB) will be 0.
Now, we can apply Raoult's law to calculate the moles of compound B needed:
P°solution = Xwater * P°water + XB * P°B
Since the vapor pressure of B is zero, the equation simplifies to:
P°solution = Xwater * P°water
We can rearrange the equation to solve for Xwater:
Xwater = P°solution / P°water
Substituting the given values:
Xwater = 23.756 torr / 42.362 torr ≈ 0.5619
Now, we can calculate the moles of B needed:
XB = 1 - Xwater = 1 - 0.5619 ≈ 0.4381 mol
Finally, we can calculate the mass of compound B needed in grams:
Mass of compound B = nB * MB
Mass of compound B = 0.4381 mol * 97.8 g/mol ≈ 42.82 g
Therefore, approximately 42 grams of compound B would need to be added to 250 grams of water to produce a solution with a vapor pressure of 23.756 torr.
To calculate the grams of compound B needed to be added to the water, we can use Raoult's law, which states that the partial pressure of a component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.
Let's denote:
P_B = vapor pressure of compound B
P_W = vapor pressure of water
X_B = mole fraction of compound B
According to Raoult's law:
P_B = X_B * P_total
Since compound B is nonvolatile, its vapor pressure can be considered negligible compared to that of water. Thus, the total vapor pressure of the solution will be the vapor pressure of water.
P_total = P_W = 42.362 torr
We want to find the grams of compound B needed to produce a solution with a vapor pressure of 23.756 torr.
P_B = 23.756 torr
Now, let's rearrange the equation to solve for X_B:
X_B = P_B / P_total
X_B = 23.756 torr / 42.362 torr
X_B = 0.5617
Since mole fraction is calculated by dividing the moles of compound B by the total moles, we can set up the following equation:
X_B = moles of B / (moles of B + moles of water)
0.5617 = moles of B / (moles of B + moles of water)
We know that moles = mass / molar mass, so we can rewrite the equation as:
0.5617 = (grams of B / molar mass of B) / [(grams of B / molar mass of B) + (grams of water / molar mass of water)]
The molar mass of water is approximately 18.02 g/mol.
Simplifying the equation:
0.5617 = (grams of B / 97.8) / [(grams of B / 97.8) + (250 / 18.02)]
Cross-multiplying:
0.5617 * [(grams of B / 97.8) + (250 / 18.02)] = grams of B / 97.8
Now we can solve for grams of B:
0.5617 * (grams of B + (250 * 97.8 / 18.02)) = grams of B
Simplifying:
0.5617 * grams of B + 0.5617 * (250 * 97.8 / 18.02) = grams of B
0.5617 * grams of B - grams of B = - 0.5617 * (250 * 97.8 / 18.02)
0.5617 * grams of B - grams of B = - 763.03
-0.4383 * grams of B = -763.03
grams of B = -763.03 / -0.4383
grams of B = 1739 (rounded to a whole number)
Therefore, approximately 1739 grams of compound B would need to be added to 250 grams of water to produce a solution with a vapor pressure of 23.756 torr.