37.8g of water to give a freezing point of -.15 degrees C

[trixie23]

Hey, I really need help in my chem homework. :(

how many grams of ethyl alcohol CH2OHCH2OH must be added to 37.8g of water to give a freezing point of -.15 degrees C.

 

[symbolipoint]

One must either find this information in a handbook or conduct a determination experimentally. I may suspect that doing the determination experimentally could be difficult, since freezing of a liquid is being watched for. Anybody have a guess, would the solution turn turbid near or at the freezing point? One would want to gently agitate the solution during cooling to uniformly distribute the temperatures.

 

[Borek]

I guess it is just about cryoscopic constant. What is cryoscopic constant for water? What molality of ethanol do you need?

 

[sjb-2812]

Once you have the molality, check you don't mean ethane-1,2-diol, (ethylene glycol), rather than ethanol - the two have different molecular weights, and so will generate different masses...

 

[DeeNos]

I can help you with your chemistry homework. To determine the amount of ethyl alcohol needed, we need to use the equation for freezing point depression, which is ΔT = Kf*m*i. In this equation, ΔT represents the change in freezing point, Kf is the freezing point depression constant, m is the molality of the solution, and i is the van't Hoff factor.

In this case, we know that the freezing point depression is -0.15 degrees C, and the molality of the solution is calculated as moles of solute (ethyl alcohol) divided by kilograms of solvent (water). We can rearrange the equation to solve for the amount of solute (ethyl alcohol) needed:

m = ΔT / (Kf * i)

First, we need to calculate the molality of the solution. We know that the molar mass of water is 18.02 g/mol and the molar mass of ethyl alcohol is 46.07 g/mol. Therefore, the moles of water in 37.8g is 37.8g / 18.02 g/mol = 2.098 moles.

Next, we need to convert the freezing point depression constant (Kf) to the appropriate units. Since the units of molality are moles/kg, we need to convert Kf from degrees C/m to degrees C*kg/mol. This can be done by multiplying Kf by the molar mass of the solvent (water), which is 18.02 g/mol. The value for Kf is typically provided in the problem, but for water it is approximately 1.86 degrees C*kg/mol.

Finally, we need to determine the van't Hoff factor (i) for ethyl alcohol. This factor takes into account the number of ions that are produced when ethyl alcohol dissolves in water. Since ethyl alcohol does not dissociate in water, the van't Hoff factor is equal to 1.

Now, we can plug in our values into the equation:

m = -0.15 degrees C / (1.86 degrees C*kg/mol * 1)

Solving for m, we get:

m = -0.15 degrees C / 1.86 degrees C*kg/mol = -0.0806 moles/kg

To determine the mass of ethyl alcohol needed, we can use the