1.)What is the boiling point of the water in your radiator if 2.00kg of antifreeze is added to 9.00 * 10^3 grams of water?
2.)Assuming that each system behaves ideally, which solution (0.25 m KCl or 0.25 m C3H8O3)would have the lower freezing point? Explain.
My answer: I say that C3H8O3 would have the lower point because of hydrogen??? right or wrong
3.)Is the following statement true or false: "Dissolving a nonvolatile solute in a solvent has the effect of extending the temperature range over which the solvent remains in the liquid phase." Explain.
My answer: I say this statement is true.????
1. You don't have enough information to work the problem.
2,000 g/molar mass of the antifreeze (whatever that happens to be) = moles antifreeze.
molality = moles/kg = moles/9 kg solvnt.
Then delta T = Kb*molality
Then added delta T to 100 C to find boiling point. (Note; this is not an accurate process to determine the boiling point of radiator water BECAUSE the system is under more than 1 atm pressure when the engine is running and that increases the boiling point, also.)
2. What in the world is C3H8O3? My guess however is that the answer is C3H8O3 but not for the reason you cite. I think it's lower because KCl breaks into two particles and the C3H8O3 PROBABLY is a covalent compound (not ionic as is KCl).
3. I think #3 is true.
its c3ho5
#3- When a non-volatile solute is added to a solvent, the resulting solution has different physical properties than the pure solvent.
1.) To find the boiling point of the water in the radiator after adding antifreeze, we need to consider the change in boiling point due to the colligative property of the solution. The boiling point elevation is given by the equation:
ΔTb = Kb * m
Where ΔTb is the change in boiling point, Kb is the molal boiling point elevation constant of the solvent (water in this case), and m is the molality of the solute (antifreeze).
First, convert the mass of water from grams to kilograms:
Water mass = 9.00 * 10^3 grams = 9.00 kilograms
Next, calculate the molality of the antifreeze:
Molality (m) = moles of solute / mass of solvent (in kg)
We need to convert the mass of antifreeze from kg to grams:
Antifreeze mass = 2.00kg = 2.00 * 10^3 grams
Next, determine the number of moles of antifreeze (using its molar mass):
Molar mass of antifreeze = (mass of antifreeze) / (moles of antifreeze)
Finally, calculate the molality:
m = (moles of antifreeze) / (mass of water)
Once you have the molality (m), you can use the boiling point elevation constant (Kb) of water to find ΔTb, the change in boiling point.
2.) To determine which solution (0.25 m KCl or 0.25 m C3H8O3) would have the lower freezing point, we need to consider the freezing point depression. The freezing point depression is given by the equation:
ΔTf = Kf * m
Where ΔTf is the change in freezing point, Kf is the molal freezing point depression constant of the solvent, and m is the molality of the solute.
In this case, both solutions have the same molality (0.25 m), so the key factor is the molal freezing point depression constant (Kf) of each solvent (KCl and C3H8O3). By comparing the Kf values for each solvent, you can determine which solution would have the lower freezing point. The solute (KCl or C3H8O3) itself does not play a role in determining the freezing point depression.
3.) The statement "Dissolving a nonvolatile solute in a solvent has the effect of extending the temperature range over which the solvent remains in the liquid phase" is false.
When a nonvolatile solute is dissolved in a solvent, it does not extend the temperature range over which the solvent remains in the liquid phase. Adding a nonvolatile solute lowers the vapor pressure of the solvent, which results in an increase in boiling point and a decrease in freezing point. This decrease in freezing point is known as freezing point depression. As a result, the solvent's temperature range over which it remains in the liquid phase is actually narrowed, not extended, when a nonvolatile solute is added.