the addition of 50 g of salt to a sample of pure water causes its freezing point to be reduced from 0.0 C to -3. 15 C. Determine how by how many degrees the freezing point dropped for every 10 g of salt added?
50g/10g = 5 cycles of 10 degrees each starting at zero C.
3.15/5 = ?
To determine how much the freezing point dropped for every 10 g of salt added, we need to calculate the freezing point depression constant, which is a property of the solvent (in this case, water). The freezing point depression constant is denoted by Kf.
The formula to calculate the freezing point depression is:
∆Tf = Kf * m
Where:
∆Tf = change in freezing point
Kf = freezing point depression constant
m = molality of the solute
In this case, we know the change in freezing point (∆Tf) and the amount of salt added (50 g). However, we need to convert the mass of salt to molality (m) before we can use the formula.
To convert the mass of salt to molality, we need the molar mass of salt. Assuming the salt is NaCl (sodium chloride):
Molar mass of NaCl = 58.44 g/mol (sodium: 22.99 g/mol, chlorine: 35.45 g/mol)
Now, let's calculate the molality:
molality (m) = moles of solute / mass of solvent (in kg)
moles of solute = mass of salt / molar mass of NaCl
= 50 g / 58.44 g/mol
≈ 0.8567 mol
Since we are dealing with water as the solvent:
mass of solvent = mass of water = 1000 g = 1 kg
Plugging the values into the formula:
∆Tf = Kf * m
∆Tf = -3.15 °C - 0.0 °C = -3.15 °C
m = 0.8567 mol / 1 kg = 0.8567 mol/kg
Now, we can rearrange the formula to solve for Kf:
Kf = ∆Tf / m
Kf = -3.15 °C / 0.8567 mol/kg
≈ -3.6799 °C/mol/kg
Now, to find the change in freezing point (∆Tf) for every 10 g of salt added, we need to calculate the molality (m) for 10 g of salt using the molar mass of NaCl as mentioned earlier. Then, we can use the formula ∆Tf = Kf * m to get the result.
moles of solute = 10 g / 58.44 g/mol
≈ 0.1713 mol
mass of solvent = 1 kg (same as before)
m = 0.1713 mol / 1 kg
≈ 0.1713 mol/kg
∆Tf = Kf * m
= -3.6799 °C/mol/kg * 0.1713 mol/kg
≈ -0.6305 °C
Therefore, the freezing point dropped by approximately 0.6305 °C for every 10 g of salt added.