lowest+freezing+point+of+Kbr+AlNO33+CH3COONa+NaNO2+MgCl2+of+same+molality
You would do well to make your post easier to read and understand.
All you need is
delta T = i*Kf*m
i = 2 for KBr.
i = 4 for Al(NO3)3
i = 2 for CH3COONa
i = 2 for NaNO2
i = 3 for MgCl2
If m is constant and Kf is constant, then delta T = some constant x i so the one with the most particles will have the lowest freezing point.
To find the lowest freezing point among the given compounds with the same molality, we need to compare their cryoscopic constants. Cryoscopic constant (Kf) is a constant specific to a solvent and represents the depression in the freezing point for a 1 molal solution of a solute.
First, let's gather the known cryoscopic constants for the given compounds:
- Kf for water (solvent) = 1.86 °C/m
Next, we need to calculate the molality (m) of each compound. Molality is the number of moles of solute dissolved in 1 kilogram of solvent.
Assuming each compound is dissolved in water:
1. KBr: Let's say the molality is 'x' mol/kg. Since KBr ionizes completely in water to give K+ and Br- ions, the number of moles of KBr is equal to 'x'. So, the molality of KBr is 'x' mol/kg.
2. Al(NO3)3: Assuming the molality is 'x' mol/kg, the number of moles of Al(NO3)3 is also equal to 'x'. Thus, the molality of Al(NO3)3 is 'x' mol/kg.
3. CH3COONa: Assuming the molality is 'x' mol/kg, the number of moles of CH3COONa is also equal to 'x'. Therefore, the molality of CH3COONa is 'x' mol/kg.
4. NaNO2: Assuming the molality is 'x' mol/kg, the number of moles of NaNO2 would be 'x'. Hence, the molality of NaNO2 is 'x' mol/kg.
5. MgCl2: Assuming the molality is 'x' mol/kg, the number of moles of MgCl2 is '2x'. So, the molality of MgCl2 is '2x' mol/kg.
Since we have an equal molality for all the compounds, we can focus only on their cryoscopic constants to determine the lowest freezing point.
Now, we will calculate the depression in the freezing point (∆Tf) for each compound using the formula: ∆Tf = Kf * m
1. For KBr: ∆Tf (KBr) = 1.86 °C/m * x mol/kg = 1.86x °C
2. For Al(NO3)3: ∆Tf (Al(NO3)3) = 1.86 °C/m * x mol/kg = 1.86x °C
3. For CH3COONa: ∆Tf (CH3COONa) = 1.86 °C/m * x mol/kg = 1.86x °C
4. For NaNO2: ∆Tf (NaNO2) = 1.86 °C/m * x mol/kg = 1.86x °C
5. For MgCl2: ∆Tf (MgCl2) = 1.86 °C/m * (2x) mol/kg = 3.72x °C
From the calculations, we can observe that the ∆Tf for MgCl2 (3.72x °C) is greater than the ∆Tf for any of the other compounds (1.86x °C). Therefore, MgCl2 will have the lowest freezing point among the given compounds with the same molality.