Using data from this table of molal boiling-point-elevation and freezing-point-depression constants, calculate the freezing and boiling points of each of the following solutions.
a)0.39 m glucose in ethanol
(b) 22.8 g of decane, C10H22, in 53.9 g CHCl3
(c) 3.90 g NaOH in 109 g of water
(d) 0.37 mol ethylene glycol and 0.27 mol KBr in 194 g H2O
how do i do these?
To solve these problems, we need to use the freezing-point-depression and boiling-point-elevation formulas:
1. Freezing-Point Depression: ΔTf = Kf × m × i
2. Boiling-Point Elevation: ΔTb = Kb × m × i
Where:
- ΔTf is the change in freezing point
- Kf is the molal freezing-point depression constant
- ΔTb is the change in boiling point
- Kb is the molal boiling-point elevation constant
- m is the molality (moles of solute per kilogram of solvent)
- i is the van't Hoff factor (the number of particles the solute dissociates into)
Let's use this information to solve each part of the problem:
a) 0.39 m glucose in ethanol:
1. Find the molality (m) of the solution:
- Molality (m) = moles of solute / kilograms of solvent
- In this case, we have 0.39 moles of glucose dissolved in ethanol. We need to find the mass of ethanol since it's given in molal units. The molar mass of glucose is 180.16 g/mol, and the density of ethanol is about 0.789 g/mL.
- Calculate the mass of ethanol: mass = volume × density
- The volume of ethanol can be found using the formula: Volume = Mass / Density
b) 22.8 g of decane, C10H22, in 53.9 g CHCl3:
1. Determine the molar mass of decane (C10H22):
- C: 12.01 g/mol × 10 = 120.1 g/mol
- H: 1.01 g/mol × 22 = 22.22 g/mol
- Sum = 120.1 g/mol + 22.22 g/mol = 142.32 g/mol
2. Calculate the number of moles of decane (C10H22) using its mass:
- Moles = mass / molar mass
3. Determine the molality (m) of the solution:
- Molality (m) = moles of solute / kilograms of solvent
- In this case, we have 22.8 g of decane and 53.9 g of CHCl3. Convert the masses to moles using the formula: Moles = Mass / Molar Mass
c) 3.90 g NaOH in 109 g of water:
1. Determine the molar mass of NaOH:
- Na: 22.99 g/mol
- O: 16.00 g/mol × 1 = 16.00 g/mol
- H: 1.01 g/mol × 1 = 1.01 g/mol
- Sum = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol
2. Calculate the number of moles of NaOH using its mass:
- Moles = Mass / Molar Mass
3. Determine the molality (m) of the solution:
- Molality (m) = moles of solute / kilograms of solvent
- In this case, we have 3.90 g of NaOH and 109 g of water. Convert the masses to moles using the formula: Moles = Mass / Molar Mass
d) 0.37 mol ethylene glycol and 0.27 mol KBr in 194 g H2O:
1. Determine the molality (m) of the solution:
- Molality (m) = moles of solute / kilograms of solvent
- In this case, we have 0.37 mol of ethylene glycol, 0.27 mol of KBr, and 194 g of water. Convert the masses to moles using the formula: Moles = Mass / Molar Mass
Once you have the molality (m) value for each solution, you can use the freezing-point-depression and boiling-point-elevation formulas mentioned earlier to calculate the freezing and boiling points of the solutions.