Two aqueous solutions of sucrose, C12H22O12, are mixed. One solution is 0.1495 M sucrose and has d = 1.018 g/mL; the other is 12.00 % sucrose by mass and has d = 1.038 g/ml.
Calculate the moles C12H22O12 percent in the mixed solution.
Please provide steps
100 mL for each soln.
A:
density = 1.018 g/mL x 100 mL = 101.8 g = mass solution.
moles = M x L = 0.1495 x 0.100 = 0.01495
g sucrose = 0.01495 x 342(molar mass but you look it up because I've rounded here and there) = 5.11 g sucrose which means the water has a mass of 101.8-5.11 = ?? g H2O.
B:
density = 1.038 g/mL x 100 mL = 103.8 g = mass solution
12.00% sucrose or 103.8 x 0.1200 = 12.46 g sucrose or 12.46/342 = 0.0364 moles sucrose.
Water has a mass of 103.8-12.46 g = ?? g.
Now take moles from A and add to moles B to obtain total moles sucrose.
Add grams water from A to grams water from B to obtain total moles H2O.
Find mole fraction sucrose (mols sucrose/(moles sucrose + moles H2O) = Xsucrose
Then mole percent = 100*Xsucrose
To calculate the moles of C12H22O12 percent in the mixed solution, we need to first find the moles of sucrose in each solution, and then add them together.
Step 1: Calculate the moles of sucrose in the first solution:
The molarity (M) of the first solution is given as 0.1495 M, which means there are 0.1495 moles of sucrose in 1 liter (1000 mL) of solution. Since we have 100 mL of solution, the moles of sucrose can be calculated using the formula:
moles = Molarity * Volume (in liters)
moles = 0.1495 mol/L * 0.1 L = 0.01495 mol
So, the first solution contains 0.01495 moles (approximately) of sucrose.
Step 2: Calculate the moles of sucrose in the second solution:
The second solution is given as 12.00% sucrose by mass. This means that for every 100 g of solution, there are 12.00 g of sucrose. Since we have 100 mL of solution (which we assume has a density of 1.038 g/mL), we can calculate the mass of the second solution using the formula:
mass = density * volume
mass = 1.038 g/mL * 0.1 L = 0.1038 g
Since the mass percentage of sucrose is 12.00%, we can calculate the mass of sucrose using the formula:
mass_sucrose = mass_solution * (mass_percentage / 100)
mass_sucrose = 0.1038 g * (0.12) = 0.01245 g
Next, we need to convert the mass of sucrose to moles by dividing by the molar mass of sucrose. The molar mass of sucrose (C12H22O12) can be calculated by adding up the atomic masses of its elements from the periodic table.
molar mass of C12H22O12 = 12(12.01 g/mol) + 22(1.01 g/mol) + 12(16.00 g/mol) = 342.3 g/mol
moles_sucrose = mass_sucrose / molar mass
moles_sucrose = 0.01245 g / 342.3 g/mol = 3.63 x 10^-5 mol
So, the second solution contains 3.63 x 10^-5 moles (approximately) of sucrose.
Step 3: Calculate the total moles of sucrose in the mixed solution:
To calculate the total moles of sucrose in the mixed solution, we simply add the moles of sucrose from each solution together.
total moles of sucrose = moles_first_solution + moles_second_solution
total moles of sucrose = 0.01495 mol + 3.63 x 10^-5 mol = 0.015285 mol
Now, we need to calculate the moles C12H22O12 percent in the mixed solution:
moles C12H22O12 percent = (total moles of sucrose / total volume of mixed solution) * 100
Assuming that the total volume of the mixed solution is 100 mL + 100 mL = 200 mL = 0.2 L:
moles C12H22O12 percent = (0.015285 mol / 0.2 L) * 100 = 7.643% (approximately)
Therefore, the moles C12H22O12 percent in the mixed solution is approximately 7.643%.