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 confuses
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.
To find the mass of ethylene glycol needed to produce a solution that freezes at -23°C, you can use the formula:
ΔT = Kf * m
where ΔT is the change in temperature (in this case, the difference between the freezing point of the pure solvent and the desired freezing point), Kf is the cryoscopic constant (for water, it is 1.86 °C/m), and m is the molality of the solution (moles of solute per kilogram of solvent).
First, let's find the molality of the solution:
Given that the density of water is 1 g/mL and we have 10.0 L of water, we can convert the volume of water to mass:
Mass of water = Volume of water * Density of water
= 10.0 L * 1 g/mL
= 10,000 g
Since 1 kg = 1000 g, the mass of water is 10 kg.
Now, we can use the freezing point depression formula:
ΔT = T°f - T°f(solvent)
where T°f is the freezing point of the pure solvent and T°f(solvent) is the desired freezing point. In this case, T°f(pure water) = 0 °C and T°f(solvent) = -23 °C. Therefore:
ΔT = 0 °C - (-23 °C)
= 23 °C
Now we can substitute the values into the formula:
23 °C = 1.86 °C/m * m
Solving for m:
m = 23 °C / 1.86 °C/m
= 12.37 m
So, the molality of the solution is 12.37 mol/kg.
Finally, to find the mass of ethylene glycol needed, we can use the formula:
m = moles of solute / mass of solvent
Rearranging the formula:
mass of solute = moles of solute / m
Since we want the molality in terms of kg, we need to convert 10 kg to 10,000 g:
mass of solute = 12.37 mol/kg * 10,000 g
= 123,700 g
Therefore, approximately 123.7 kg of ethylene glycol must be added to 10.0 L of water to produce a solution for use in a car's radiator that freezes at -23 °C.
To find the amount of ethylene glycol needed to produce a solution that freezes at -23°C, you can use the formula for freezing point depression:
ΔT = Kf * m
where:
ΔT is the change in freezing point
Kf is the molal freezing point depression constant
m is the molality of the solution (moles of solute per kilogram of solvent)
First, let's calculate the moles of water in 10.0 L of water using its density. Given that the density is 1 g/mL, and 1 L is equal to 1000 mL, the mass of water is:
mass of water = volume × density = 10.0 L * 1000 mL/L * 1g/mL = 10,000 g
Next, we convert the mass of water to moles of water using its molar mass. The molar mass of water (H2O) is approximately 18.015 g/mol:
moles of water = mass of water / molar mass of water = 10,000 g / 18.015 g/mol = 555.42 mol
Now, let's calculate the molality of the solution. Since the mass of the solvent (water) is equal to the mass of the solution, we can use the mass of water to calculate the molality:
mass of solution = mass of water = 10,000 g
moles of solute = moles of ethylene glycol (C2H6O2)
molality = moles of solute / mass of solvent (in kg) = moles of ethylene glycol / 10.0 kg (since 10.0 L water is approximately 10.0 kg)
To solve for moles of ethylene glycol, rearrange the equation to:
moles of ethylene glycol = molality * mass of solvent (in kg)
Now we can substitute the values and find the number of moles of ethylene glycol:
moles of ethylene glycol = molality * mass of solvent (in kg)
= molality * 10.0 kg
To find the molality (m), we need to use the given change in freezing point (ΔT) of -23°C and the molal freezing point depression constant (Kf) for water. For water, the Kf value is approximately 1.86 °C/m.
We can substitute the values and find the molality (m):
ΔT = Kf * m
-23°C = 1.86°C/m * m
Simplifying the equation:
-23°C = 1.86°C * m
m = -23°C / 1.86°C
m ≈ -12.37
Since molality cannot be negative, we ignore the negative sign and use m ≈ 12.37.
Finally, substitute the calculated molality and the mass of the solvent into the equation to solve for the moles of ethylene glycol:
moles of ethylene glycol = 12.37 * 10.0 kg ≈ 123.7 mol
To convert the moles of ethylene glycol to grams, we use its molar mass. The molar mass of ethylene glycol (C2H6O2) is approximately 62.068 g/mol:
mass of ethylene glycol = moles of ethylene glycol * molar mass of ethylene glycol
≈ 123.7 mol * 62.068 g/mol
≈ 7,674.08 g
Therefore, you would need approximately 7,674.08 grams (or 7.67 kg) of ethylene glycol to produce a solution with a freezing point of -23°C when mixed with 10.0 L of water.