how much ethyl alcohol must be added to 1.00l of water so that the solution will not freeze at -4f
Answered above.
To determine the amount of ethyl alcohol that needs to be added to 1.00 liter of water in order to prevent it from freezing at -4°F, we need to calculate the freezing point depression using the formula:
ΔT = Kf * m
Where:
ΔT is the change in freezing point
Kf is the cryoscopic constant (freezing point depression constant) for the solvent
m is the molality of the solute (ethyl alcohol in this case)
The cryoscopic constant for water is approximately 1.86 °C*kg/mol. We'll need to convert -4°F to Celsius before plugging it into the equation.
Converting -4°F to Celsius:
-4°F = -20°C
Now we can calculate the molality (m) of the solute:
molality = moles of solute / mass of solvent (in kg)
Since we have 1.00 liter of water, which has a density of 1 kg/L, the mass is 1 kg.
Let's assume that the molar mass of ethyl alcohol (C2H6O) is approximately 46.07 g/mol.
Now we calculate the moles of ethyl alcohol needed. We need to find the ΔT required to prevent freezing at -4°C:
ΔT = (-20°C - 0°C) = -20°C
Now let's calculate the molality:
m = moles of ethyl alcohol / kg of water
Rearranging the equation, we get:
moles of ethyl alcohol = m * kg of water
Now let's substitute the values and solve for moles of ethyl alcohol:
moles of ethyl alcohol = m * 1 kg
Finally, we use the molar mass of ethyl alcohol to convert moles to grams or milliliters, depending on the concentration unit you prefer.
Please note that the calculations are approximate, and actual values may vary slightly due to factors like impurities and deviations from ideal behavior.