Calculate the new freezing point and the new boiling point of an aqueous solution that contains 40.5 g of Calcium Nitrate dissolved in 225 grams of water.
To calculate the new freezing point and boiling point of the solution, we need to use the concept of colligative properties. In this case, we will be using the formula for calculating the change in freezing point and boiling point, known as the van't Hoff factor.
The van't Hoff factor (i) represents the number of particles into which a solute dissociates in a solution. For calcium nitrate (Ca(NO3)2), it dissociates into three particles: one calcium ion (Ca2+) and two nitrate ions (NO3-).
Now, let's calculate the molality of the calcium nitrate solution using the formula:
Molality (m) = moles of solute / mass of solvent (in kg)
1. Calculate the moles of calcium nitrate:
Mass of calcium nitrate = 40.5 g
Molar mass of calcium nitrate (Ca(NO3)2) = 40.08 g/mol (Ca: 40.08 g/mol, N: 14.01 g/mol, O: 16.00 g/mol)
Moles of calcium nitrate = mass / molar mass = 40.5 g / 164.09 g/mol = 0.2469 mol
2. Calculate the mass of the water in kilograms:
Mass of water = 225 g = 0.225 kg
3. Calculate the molality:
Molality (m) = 0.2469 mol / 0.225 kg = 1.096 M
Now, we can use the equations to calculate the change in freezing point and boiling point of the water due to the presence of the solute.
Change in freezing point (∆Tf) = -Kf * m
Change in boiling point (∆Tb) = Kb * m
Kf and Kb are the cryoscopic and ebullioscopic constants, respectively, which are specific to the solvent. For water, their values are:
Kf = 1.86 °C·kg/mol
Kb = 0.512 °C·kg/mol
Substituting the values in the equations:
∆Tf = -1.86 °C·kg/mol * 1.096 mol/kg = -2.0376 °C
∆Tb = 0.512 °C·kg/mol * 1.096 mol/kg = 0.5619 °C
Finally, we can calculate the new freezing point and boiling point by adding the changes to the normal freezing and boiling points of water.
New Freezing Point = Normal Freezing Point (0 °C) + ∆Tf = 0 °C - 2.0376 °C = -2.0376 °C
New Boiling Point = Normal Boiling Point (100 °C) + ∆Tb = 100 °C + 0.5619 °C = 100.5619 °C
Therefore, the new freezing point of the solution is -2.0376 °C, and the new boiling point is 100.5619 °C.