What are the boiling point and freezing point of a 0.22 m solution of sucrose in ethanol?

Can someone help me with this quickly? I'm SO confused~!

delta T = Kb*molality.

Look up Kb for ethanol and solve for delta T.
You know the normal boiling point of ethanol, simply add delta T to find the new boiling point.

Same procedure for freezing point except you subtract delta T from 0 C to find the new freezing point.
delta T = Kf*m
You know the normal

To find the boiling point and freezing point of a solution, we need to consider the concept of colligative properties. Colligative properties depend on the number of solute particles dissolved in a solvent, rather than on the nature of the solute particles themselves.

First, we need to know the molality of the solution, which is the moles of solute per kilogram of solvent. In this case, the molality (m) of the solution is given as 0.22 m.

Now, we'll use the formulas for boiling point elevation and freezing point depression to calculate the changes in boiling point and freezing point, respectively.

The boiling point elevation (ΔTb) can be calculated using the formula:

ΔTb = Kb * m

Where Kb is the molal boiling point elevation constant (for ethanol, it is approximately 1.2 °C/m). Substituting the given values, we have:

ΔTb = 1.2 °C/m * 0.22 m = 0.264 °C

The boiling point of pure ethanol is approximately 78.3 °C. Therefore, the boiling point of the solution can be found by adding the boiling point elevation to the boiling point of pure ethanol:

Boiling point of solution = 78.3 °C + 0.264 °C = 78.564 °C

Next, the freezing point depression (ΔTf) can be calculated using the formula:

ΔTf = Kf * m

Where Kf is the molal freezing point depression constant (for ethanol, it is approximately 1.9 °C/m). Substituting the given values, we have:

ΔTf = 1.9 °C/m * 0.22 m = 0.418 °C

The freezing point of pure ethanol is approximately -114.1 °C. Therefore, the freezing point of the solution can be found by subtracting the freezing point depression from the freezing point of pure ethanol:

Freezing point of solution = -114.1 °C - 0.418 °C = -114.518 °C

Therefore, the boiling point of a 0.22 m solution of sucrose in ethanol is approximately 78.564 °C, and the freezing point is approximately -114.518 °C.

Sure, I can help you with that! To determine the boiling point and freezing point of a solution, you need to consider the colligative properties of the solute (sucrose) and solvent (ethanol). Colligative properties depend on the number of particles present in the solution, regardless of their identity.

Here's how you can determine the boiling point and freezing point of the solution:

1. Determine the molality (m) of the sucrose solution. Molality is defined as the moles of solute per kilogram of solvent. In this case, the molality is given as 0.22 m, which means there are 0.22 moles of sucrose in 1 kilogram of ethanol.

2. Calculate the boiling point elevation using the equation: ΔTb = Kb * m. ΔTb is the boiling point elevation, Kb is the boiling point constant for the solvent, and m is the molality. The value of Kb for ethanol is approximately 1.99 °C/m. Plug in the values to find ΔTb.

3. Add the boiling point elevation to the normal boiling point of pure ethanol (78.4 °C) to find the boiling point of the solution. The normal boiling point is the temperature at which a pure liquid boils at a given pressure.

4. Calculate the freezing point depression using the equation: ΔTf = Kf * m. ΔTf is the freezing point depression, Kf is the freezing point constant for the solvent, and m is the molality. The value of Kf for ethanol is approximately 1.86 °C/m. Plug in the values to find ΔTf.

5. Subtract the freezing point depression from the normal freezing point of pure ethanol (-114.1 °C) to find the freezing point of the solution. The normal freezing point is the temperature at which a pure liquid freezes at a given pressure.

By following these steps, you can determine the boiling point and freezing point of the 0.22 m solution of sucrose in ethanol.