A solution contains .370 mol of isopropanol (C3H7OH) dissolved in 1.700 mol of water. What is the mass percent of isopropanol? What is the molality of isopropanol? i've done it a couple times now and i must have something off
I answered this question for Natalie earlier. If you are not Natalie, here is a link to that question and my response.
http://www.jiskha.com/display.cgi?id=1287432814
To find the mass percent of isopropanol in the solution, you need to calculate the mass of isopropanol and the total mass of the solution.
1. Calculate the mass of isopropanol:
To do this, you need to know the molar mass of isopropanol (C3H7OH). Let's calculate it:
Molar mass of C = 12.01 g/mol
Molar mass of H = 1.008 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of isopropanol (C3H7OH) = (3 × 12.01 g/mol) + (8 × 1.008 g/mol) + (1 × 16.00 g/mol)
= 60.10 g/mol
Mass of isopropanol = Number of moles of isopropanol × Molar mass of isopropanol
= 0.370 mol × 60.10 g/mol
= 22.207 g
2. Calculate the total mass of the solution:
The total mass of the solution is the sum of the mass of isopropanol and the mass of water:
Mass of solution = Mass of isopropanol + Mass of water
= 22.207 g + (1.700 mol × 18.02 g/mol)
= 22.207 g + 30.634 g
= 52.841 g
3. Calculate the mass percent of isopropanol:
Mass percent of isopropanol = (Mass of isopropanol / Mass of solution) × 100
= (22.207 g / 52.841 g) × 100
≈ 41.97%
Now, let's calculate the molality of isopropanol.
Molality (m) is defined as the number of moles of solute per kilogram of solvent.
4. Calculate the mass of water:
Mass of water = Number of moles of water × Molar mass of water
= 1.700 mol × 18.02 g/mol
= 30.634 g
5. Calculate the molality of isopropanol:
Molality (m) = Number of moles of isopropanol / Mass of water (in kg)
= 0.370 mol / (30.634 g / 1000)
= 12.08 mol/kg
Therefore, the mass percent of isopropanol in the solution is approximately 41.97%, and the molality of isopropanol is approximately 12.08 mol/kg.
To find the mass percent of isopropanol, we need to calculate the mass of isopropanol in the solution and divide it by the total mass of the solution, then multiply by 100.
1. First, we calculate the mass of isopropanol in the solution. To do this, we need to know the molecular weight (or molar mass) of isopropanol. The molecular weight of isopropanol (C3H7OH) can be calculated by adding up the atomic weights of each individual atom:
C (Carbon) : 12.01 g/mol
H (Hydrogen) : 1.01 g/mol
O (Oxygen) : 16.00 g/mol
Molecular weight of isopropanol (C3H7OH) = 12.01 * 3 + 1.01 * 7 + 16.00 * 1 = 60.10 g/mol
2. Next, we calculate the mass of isopropanol in the solution:
Mass of isopropanol = moles of isopropanol * molecular weight of isopropanol
Given: Moles of isopropanol = 0.370 mol
Mass of isopropanol = 0.370 mol * 60.10 g/mol = 22.207 g
3. Now, we calculate the total mass of the solution by adding the mass of isopropanol to the mass of water:
Total mass of solution = mass of isopropanol + mass of water
Given: Mass of water = 1.700 mol (Assuming the density of water is approximately 1 g/mL)
Mass of water = 1.700 mol * 18.02 g/mol = 30.634 g
Total mass of solution = 22.207 g + 30.634 g = 52.841 g
4. Finally, we calculate the mass percent of isopropanol:
Mass percent of isopropanol = (mass of isopropanol / total mass of solution) * 100
Mass percent of isopropanol = (22.207 g / 52.841 g) * 100 = 41.97%
Now, let's calculate the molality of isopropanol.
Molality is defined as the moles of solute (isopropanol) per kilogram of solvent (water).
Given: Moles of isopropanol = 0.370 mol
Mass of water = 1.700 mol * 18.02 g/mol = 30.634 g
Density of water ≈ 1 g/mL ≈ 1000 g/L
1. First, we convert the mass of water to kilograms:
Mass of water (in kg) = 30.634 g / 1000 = 0.030634 kg
2. Now, we calculate the molality of isopropanol:
Molality = moles of isopropanol / mass of water (in kg)
Molality = 0.370 mol / 0.030634 kg ≈ 12.067 mol/kg
Therefore, the mass percent of isopropanol is approximately 41.97%, and the molality of isopropanol is approximately 12.067 mol/kg.