How much ethylalcohol must be added to 1litre of water so that the solution will freeze at 14 farenite, kf of wa
ter is 1.86 c/mol
To find the amount of ethyl alcohol that must be added to 1 liter of water, we need to use the freezing point depression equation:
ΔT = Kf * m,
where:
ΔT is the freezing point depression,
Kf is the molal freezing point depression constant for water (1.86 °C/mol),
m is the molality of the solute (ethyl alcohol).
Before we can calculate the molality, we need to convert 14 Fahrenheit to Celsius:
T(°C) = (T(°F) - 32) * 5/9,
T(°C) = (14 - 32) * 5/9,
T(°C) = -7.78.
Now, let's calculate the molality:
molality (m) = moles of solute (ethyl alcohol) / mass of the solvent (water in kg).
Since we are adding ethyl alcohol to water, the mass of the solvent (water) is 1 kg (since 1 liter of water weighs approximately 1 kg).
Let's assume x is the amount of ethyl alcohol added in moles.
So, the molality (m) can be calculated as:
m = x mol / 1 kg,
m = x mol.
Now, we can substitute the values into the freezing point depression equation:
ΔT = Kf * m,
-7.78 = 1.86 * x.
Solving for x:
x = -7.78 / 1.86,
x ≈ -4.18 moles.
Since moles cannot be negative, we cannot have negative quantities of ethyl alcohol. Therefore, there is no amount of ethyl alcohol that can be added to 1 liter of water to freeze the solution at 14 Fahrenheit.
To calculate the amount of ethyl alcohol that must be added to 1 liter of water so that the solution will freeze at 14 degrees Fahrenheit, we'll need to use the freezing point depression equation and the given constants.
The freezing point depression equation is:
ΔT = Kf * m
Where:
ΔT is the change in temperature (in Celsius)
Kf is the molal freezing point depression constant for the solvent (in Celsius/mole)
m is the molality of the solute (in moles of solute per kilogram of solvent)
Given:
The freezing point depression constant (Kf) of water = 1.86 °C/m
We need to find the amount of ethyl alcohol in moles (mol) needed to achieve a freezing point depression of 14 °F (-10 °C).
Step 1: Convert the freezing point depression to Celsius.
14 °F = -10 °C
Step 2: Calculate the change in temperature (ΔT) in Celsius.
ΔT = -10 °C
Step 3: Convert liters to kilograms.
Since the density of water is 1 g/mL, 1 liter of water weighs 1000 grams or 1 kilogram.
Step 4: Calculate the molality (m) of the ethyl alcohol.
m = moles of solute / kilograms of solvent
Step 5: Rearrange the freezing point depression equation to solve for moles of solute.
moles of solute = ΔT / Kf * m
Step 6: Plug in the values and solve for moles of solute (ethyl alcohol).
moles of solute = -10 °C / (1.86 °C/mol) * (1 kg/1000 g)
moles of solute = -10/1.86 * 0.001
moles of solute ≈ -0.00538 mol
Since we cannot have a negative amount of ethyl alcohol, it is not possible to add a negative amount to the water. Therefore, ethyl alcohol cannot be added to the water to achieve a freezing point of 14 degrees Fahrenheit.
14 F. Convert to C. (F-32)*5/9 = C
delta T = Kf*molality
Substitute delta T and Kf. Solve for m.
m = mols/kg solvent.
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