Which would be more effective in lowering the freezing point of 500 g of water? 100 g of sucrose or 100 g of ethyl alcohol?

molality = moles/kg solvent

moles sucrose = 100/molar mass sucrose.
moles ethanol = 100/molar mass ethanol.

delta T = Kf*molality

To determine which substance would be more effective in lowering the freezing point of water, we need to compare their respective freezing point depression constants.

The freezing point depression constant (Kf) is a property of each substance and indicates how much the freezing point of a solution will decrease for every mole of solute dissolved in a solvent.

For sucrose:
1. Convert the given mass of sucrose to moles using its molar mass.
The molar mass of sucrose (C12H22O11) is approximately 342.3 g/mol.

Mass of sucrose = 100 g
Moles of sucrose = 100 g ÷ 342.3 g/mol = 0.2922 mol

2. Find the freezing point depression constant (Kf) for sucrose.
The Kf value for sucrose is approximately 1.86 °C/m.

3. Calculate the freezing point depression caused by 0.2922 mol of sucrose in 500 g of water.
ΔT = Kf × molality
Molality (m) = moles of solute ÷ mass of solvent (in kg)

Mass of water = 500 g = 0.5 kg
Molality = 0.2922 mol ÷ 0.5 kg = 0.5844 mol/kg

ΔT = 1.86 °C/m × 0.5844 mol/kg = 1.0874 °C

For ethyl alcohol (ethanol):
1. Convert the given mass of ethyl alcohol to moles using its molar mass.
The molar mass of ethyl alcohol (C2H5OH) is approximately 46.07 g/mol.

Mass of ethyl alcohol = 100 g
Moles of ethyl alcohol = 100 g ÷ 46.07 g/mol = 2.1719 mol

2. Find the freezing point depression constant (Kf) for ethyl alcohol.
The Kf value for ethyl alcohol is approximately 1.86 °C/m (same as sucrose).

3. Calculate the freezing point depression caused by 2.1719 mol of ethyl alcohol in 500 g of water.
ΔT = Kf × molality
Molality (m) = moles of solute ÷ mass of solvent (in kg)

ΔT = 1.86 °C/m × (2.1719 mol ÷ 0.5 kg) = 8.0106 °C

Comparing the two calculations, we see that the freezing point depression caused by 100 g of ethyl alcohol is larger than that caused by 100 g of sucrose. Therefore, in this case, 100 g of ethyl alcohol would be more effective in lowering the freezing point of 500 g of water.

To determine which substance would be more effective in lowering the freezing point of water, we need to compare their respective freezing point depression constants. The freezing point depression constant, also known as the cryoscopic constant, is a characteristic property of a substance that quantifies its ability to decrease the freezing point of a solvent.

The equation to calculate the freezing point depression is:

ΔT = K * m

where:
ΔT is the change in freezing point,
K is the cryoscopic constant, and
m is the molality of the solute.

To find the cryoscopic constant (K), we need to know the molality (m) of the solute. Molality is defined as the number of moles of solute dissolved in 1 kilogram of solvent.

Let's calculate the molality (m) first:

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

For 100 g of a substance:
Number of moles of sucrose (C12H22O11) = mass / molar mass
Number of moles of ethyl alcohol (C2H5OH) = mass / molar mass

Molar mass of sucrose = 342.3 g/mol
Molar mass of ethyl alcohol = 46.07 g/mol

Now, calculate the molality (m) for both substances.

For sucrose:
m = (100 g) / (0.5 kg) = 200 mol/kg

For ethyl alcohol:
m = (100 g) / (0.5 kg) = 200 mol/kg

Notice that the molalities for both substances are the same since the mass of solute and solvent is constant.

The cryoscopic constant (K) for sucrose is 1.86 °C·kg/mol, and for ethyl alcohol, it is 1.99 °C·kg/mol.

Now, substitute the values into the equation to compare the change in freezing point (ΔT) for each substance:

ΔT(sucrose) = 1.86 °C·kg/mol * 200 mol/kg = 372 °C
ΔT(ethyl alcohol) = 1.99 °C·kg/mol * 200 mol/kg = 398 °C

Comparing the ΔT values, we find that 100 g of ethyl alcohol would be more effective in lowering the freezing point of 500 g of water than 100 g of sucrose.