Answer & Explanation:              PA BRAINLIEST

1. To solve this problem, we can use the formula for vapor pressure lowering:

ΔP = X2 * P°2

where X2 is the mole fraction of the solute (glycerol), P°2 is the vapor pressure of the pure solvent (water), and ΔP is the vapor pressure lowering.

First, we need to calculate the number of moles of glycerol and water in the solution:

moles of glycerol = mass of glycerol / molar mass of glycerol

= 10.0 mL * 1.26 g/mL / 92.09 g/mol

= 0.1372 mol

moles of water = mass of water / molar mass of water

= 500.0 mL * 0.988 g/mL / 18.015 g/mol

= 27.533 mol

The total moles of the solution is the sum of the moles of glycerol and water:

total moles = moles of glycerol + moles of water

= 0.1372 mol + 27.533 mol

= 27.6702 mol

Next, we can calculate the mole fraction of glycerol:

X2 = moles of glycerol / total moles

= 0.1372 mol / 27.6702 mol

= 0.00496

Finally, we can calculate the vapor pressure lowering:

ΔP = X2 * P°2

= 0.00496 * 92.5 torr

= 0.4592 torr

Therefore, the vapor pressure lowering when 10.0 mL of glycerol is added to 500.0 mL of water at 50.oC is 0.4592 torr.

2. To calculate the boiling and freezing points of the solution, we first need to calculate the molality (m) of the solution, which is defined as the number of moles of solute per kilogram of solvent.

The molar mass of ethylene glycol (C2H6O2) is:

Molar mass of C = 12.01 g/mol

Molar mass of H = 1.01 g/mol

Molar mass of O = 16.00 g/mol

Total molar mass = 2(12.01) + 6(1.01) + 2(16.00) = 62.07 g/mol

Number of moles of ethylene glycol = mass / molar mass

= 3000 g / 62.07 g/mol

= 48.36 mol

Mass of water = 2500 g

Molality of the solution = moles of solute / mass of solvent (in kg)

= 48.36 mol / 2.5 kg

= 19.344 mol/kg

Now we can use the following equations to calculate the boiling and freezing points of the solution:

ΔTb = Kb × m

ΔTf = Kf × m

where Kb and Kf are the molal boiling point elevation and freezing point depression constants for water, respectively.

Substituting the given values, we get:

ΔTb = 0.512 oC/m × 19.344 mol/kg = 9.90 oC

ΔTf = 1.86 oC/m × 19.344 mol/kg = 35.95 oC

Therefore, the boiling point of the solution is 100 + 9.90 = 109.90 oC and the freezing point is 0 - 35.95 = -35.95 oC.

3. The freezing point depression constant (Kf) for water is 1.86 °C/m. To calculate the minimum concentration of ethylene glycol solution that will protect the cooling system from freezing at 0.0°C, we can use the equation:

ΔTf = Kf x molality

where ΔTf is the freezing point depression, Kf is the freezing point depression constant, and molality is the concentration of the solution in mol solute per kg solvent.

At the freezing point, ΔTf = 0, so we can rearrange the equation to solve for the molality:

molality = 0 / Kf

molality = 0 / 1.86 = 0 mol/kg

This means that in order to prevent freezing at 0.0°C, we need a minimum concentration of 0 mol/kg, or in other words, pure ethylene glycol would be required. However, it is important to note that a higher concentration of ethylene glycol is typically used in automotive cooling systems to provide a greater margin of safety.

4. (a) To calculate the molality of the solution, we first need to calculate the freezing point depression (ΔTf) of the solution:

ΔTf = Kf × molality

where Kf is the freezing point depression constant of water (1.86 °C/m) and molality is the molality of the solution.

We know that the freezing point depression of the solution is -0.450 °C, so we can rearrange the equation to solve for molality:

molality = ΔTf / Kf

molality = (-0.450 °C) / (1.86 °C/m)

molality = -0.242 mol/kg (Note: the negative sign indicates a decrease in temperature)

Therefore, the molality of the nicotine solution is -0.242 mol/kg.

(b)To calculate the molar mass of nicotine, we first need to calculate the number of moles of nicotine in the solution. We can use the molality calculated in part (a) to do so:

molality (m) = moles of solute / mass of solvent (in kg)

Rearranging this equation, we get:

moles of solute = molality x mass of solvent (in kg)

mass of solvent = 48.92 g = 0.04892 kg

moles of solute = (0.04625 mol/kg) x 0.04892 kg = 0.002261 mol

Now, we can calculate the molar mass of nicotine using the following equation:

molar mass = mass / moles

mass of nicotine = 1.921 g = 0.001921 kg

molar mass = 0.001921 kg / 0.002261 mol = 848.7 g/mol

Therefore, the molar mass of nicotine is approximately 848.7 g/mol.