Question

31.)Show the calculation of the boiling point of a solution made by dissolving 15.9 grams of the nonelectrolyte C₃H₇OH in 350 grams of water. K_b for water is 0.51, BP of pure water is 100^oC.

30.)Show the calculation of the freezing point of a solution made by dissolving 12.6 grams of the nonelectrolyte C₂H₅OH in 200 grams of water. K_f for water is 1.86, FP of pure water is 0^oC.

29.)Show the calculation of the volume of 0.238 M solution which can be prepared using 13.4 grams of Ca₃(PO₄)₂.

28.)Show the calculation of the volume of 0.987 M solution which can be prepared using 24.6 grams of NaNO₃.

Answer 1

31. To calculate the boiling point elevation, we use the formula:

ΔTb = Kb * molality

First, we need to calculate the molality of the solution:

molality = moles of solute / mass of solvent in kg

We have 15.9 g of C3H7OH, whose molar mass is 60.1 g/mol. Thus, we have:

moles of C3H7OH = 15.9 g / 60.1 g/mol = 0.265 mol

We have 350 g of water, which is 0.350 kg. Thus, we have:

molality = 0.265 mol / 0.350 kg = 0.757 mol/kg

Now we can calculate the boiling point elevation:

ΔTb = 0.51 * 0.757 = 0.386°C

Therefore, the boiling point of the solution is:

Boiling point = 100°C + 0.386°C = 100.386°C

28. To calculate the volume of solution, we use the formula:

moles of solute = Molarity * volume of solution in liters

First, we need to calculate the moles of NaNO3:

moles of NaNO3 = 24.6 g / 85.0 g/mol = 0.289 mol

Now we can calculate the volume of solution:

volume of solution = moles of solute / Molarity

Molarity = 0.987 mol/L

volume of solution = 0.289 mol / 0.987 mol/L = 0.293 L

Therefore, the volume of solution is 0.293 L or 293 mL.

29. To calculate the volume of solution, we use the formula:

moles of solute = Molarity * volume of solution in liters

First, we need to calculate the moles of Ca3(PO4)2:

moles of Ca3(PO4)2 = 13.4 g / (340.1 g/mol + 230.97 g/mol + 8*16.0 g/mol) = 0.0348 mol

Now we can calculate the volume of solution:

volume of solution = moles of solute / Molarity

Molarity = 0.238 mol/L

volume of solution = 0.0348 mol / 0.238 mol/L = 0.146 L

Therefore, the volume of solution is 0.146 L or 146 mL.

30. To calculate the freezing point depression, we use the formula:

ΔTf = Kf * molality

First, we need to calculate the molality of the solution:

molality = moles of solute / mass of solvent in kg

We have 12.6 g of C2H5OH, whose molar mass is 46.1 g/mol. Thus, we have:

moles of C2H5OH = 12.6 g / 46.1 g/mol = 0.273 mol

We have 200 g of water, which is 0.200 kg. Thus, we have:

molality = 0.273 mol / 0.200 kg = 1.365 mol/kg

Now we can calculate the freezing point depression:

ΔTf = 1.86 * 1.365 = 2.54°C

Therefore, the freezing point of the solution is:

Freezing point = 0°C - 2.54°C = -2.54°C