Whoever you are --
Please DO NOT USE TUTORS' NAMES where your name or nickname should be.
~~~~~~~~~~~~~~~~~~~~~~~~~~
as i earlier posted and , 1 day ago , i am confusesd hw we ll find freezing point......isnt it T pure solvent - T solution .. which ll be 0-(-23.3) = 23.3
plz conform .. and do we have to use Kf.m .. formula in order to get
( -23.3)(1)= -23.3 .. as someone earlier posted plz can u conform thnks
i am confused
What mass of ethylene glycol C2H6O2, the main component of antifreeze, must be added to 10.0 L water to produce a solution for use in a car's radiator that freezes at -23C? Assume density for water is exactly 1g/mL.
Note: It would be much much easier to read your question if you wrote it in ordinary English which means no text type words, end sentence with a period and start a new sentence with a capital letter.
Now for the question.
delta T = Kf*m (yes, Kf*m must be used).
You know delta T. It is 23 from the problem (0 - (-23) = 23.
Kf = 1.86. Use the formula to calculate m, molality.
Then use m = #mols/kg solvent.
You know m from the previous problem and you know kg solvent (it is 10.0L with a density of 1 g/mL), calculate mols.
Now use mols = grams/molar mass. You know mols and molar mass which allows you to calculate grams.
To find the mass of ethylene glycol required to produce a solution that freezes at -23°C, we can use the formula for freezing point depression:
∆T = Kf * m
Where:
∆T is the change in freezing point (in this case, -23°C)
Kf is the cryoscopic constant (given for the solvent, usually in °C/m or K/m)
m is the molality of the solute in the solution (mol solute/kg solvent)
In this case, we need to find the mass of ethylene glycol (C2H6O2) that should be added to 10.0 L of water to produce a solution that freezes at -23°C.
First, we need to convert the volume of water (10.0 L) to mass using its density. Since the density of water is given as exactly 1 g/mL, the mass of 10.0 L of water is 10.0 kg.
Next, we can calculate the molality (m) of the ethylene glycol in the solution. Molality is defined as the moles of solute divided by the mass of the solvent (in kg).
To find the moles of ethylene glycol, we need to determine the molar mass of C2H6O2:
C: 12.01 g/mol
H: 1.01 g/mol
O: 16.00 g/mol
Molar mass of C2H6O2 = (2 * 12.01 g/mol) + (6 * 1.01 g/mol) + (2 * 16.00 g/mol) = 62.07 g/mol
Now we can find the moles of ethylene glycol in 10.0 kg of water by dividing the mass (in grams) by the molar mass:
moles = mass / molar mass
moles = 10.0 kg * (1000 g/kg) / 62.07 g/mol
The resulting value is the moles of ethylene glycol in our solution.
Finally, we can calculate the molality by dividing the moles of ethylene glycol by the mass of water (in kg):
m = moles / mass of water
Now that we know ∆T, Kf, and m, we can rearrange the formula to solve for the freezing point depression:
∆T = Kf * m
-23°C = Kf * m
Now, you can use the given freezing point (∆T), Kf value for ethylene glycol-water solution, and the obtained molality (m) to solve for the missing variable, which is the Kf value in your case.
Once you have the Kf value, you can substitute it into the equation to find the mass of ethylene glycol (molality * mass of water) needed to produce the desired freezing point.