how many grams of CaCl2 need to be dssolved in 100g of water to prepare a solution which freezes at -2.0 degrees?
Work this problem just like the previous one EXCEPT there is another part to the formula.
Delta T = i*Kf *m
i is the number of particles and CaCl2 will have 3. In the previous problem i=1 and I omitted it from the equation.
in this equation is Kf equal you 2 ??
okay so delta t= 2 i= 3 when you find moles do you turn them into grams and that is the answer??
Kf = 1.86o C/m in both problems. That is the freezing point constant for water. Solve for molality, and turn that into grams in 100 g water. Post your work if you want me to check it.
To solve this problem, we need to use the formula for freezing point depression:
ΔT = i * Kf * m
Where:
ΔT is the change in temperature (in degrees Celsius)
i is the van 't Hoff factor, a measure of the number of particles the solute dissociates into in solution (in this case, CaCl2 dissociates into 3 ions, so i = 3)
Kf is the freezing point depression constant for the solvent (in this case, water - Kf = 1.86 °C/m)
m is the molality of the solution, which is the ratio of the moles of solute to the mass of the solvent (in this case, water)
In this problem, you are given that the solution freezes at -2.0 degrees Celsius. So, we can plug in the given values into the formula:
-2.0 = 3 * 1.86 * m
Now, we can solve for the molality (m):
m = -2.0 / (3 * 1.86)
m ≈ -0.358 mol/kg
Note: The negative sign indicates that the solution is depressed below the freezing point of pure water.
To convert molality into grams of CaCl2, you need to know the molar mass of CaCl2, which is approximately 110.98 g/mol.
First, calculate the number of moles of CaCl2:
moles of CaCl2 = m * (mass of water in kg)
Since you're given 100 g of water, the mass of water in kg is 0.1 kg. So:
moles of CaCl2 = -0.358 * 0.1
Now, convert the moles of CaCl2 into grams:
grams of CaCl2 = moles of CaCl2 * molar mass of CaCl2
grams of CaCl2 = -0.358 * 0.1 * 110.98
grams of CaCl2 ≈ -3.97 g
Note: The negative sign again indicates that the mass of CaCl2 required is actually lower than the calculated value. However, we cannot have a negative mass, so we should take the absolute value:
grams of CaCl2 ≈ 3.97 g
Therefore, approximately 3.97 grams of CaCl2 need to be dissolved in 100 grams of water to prepare a solution that freezes at -2.0 degrees Celsius.
To solve for the number of grams of CaCl2 needed to be dissolved in 100g of water to prepare a solution that freezes at -2.0 degrees Celsius, we can use the equation:
ΔT = i*Kf*m
In this equation, ΔT represents the change in freezing point, i represents the number of particles (in this case, 3 for CaCl2), Kf represents the freezing point constant for water (which is 1.86 °C/m), and m represents the molality of the solution.
Since we want the solution to freeze at -2.0 degrees Celsius, the change in freezing point, ΔT, would be:
ΔT = -2.0 °C - 0 °C = -2.0 °C
Substituting the given values into the equation:
-2.0 °C = 3 * (1.86 °C/m) * m
Now, we can solve for m (molality):
m = -2.0 °C / (3 * 1.86 °C/m) = -2.0 °C / 5.58 °C/m ≈ -0.358 m
Since the molality (m) is negative, it implies that we have an excess amount of solute (CaCl2) for the given amount of water.
To convert molality into grams, we can use the formula:
m = moles of solute / mass of solvent (in kg)
Since there are 100g of water (0.1 kg) in the solution:
-0.358 m = moles of CaCl2 / 0.1 kg
Rearranging the equation:
moles of CaCl2 = -0.358 m * 0.1 kg = -0.0358 mol
Finally, to find the number of grams of CaCl2 needed, multiply the moles of CaCl2 by its molar mass (110.98 g/mol):
grams of CaCl2 = -0.0358 mol * (110.98 g/mol) ≈ -3.97 g
Note that the negative sign suggests that you have an excess amount of CaCl2 for the given solution. However, it is not physically meaningful to have negative grams of a substance, so the actual mass of CaCl2 required should be rounded up to zero.
Therefore, approximately 0 grams of CaCl2 should be dissolved in 100g of water to prepare a solution that freezes at -2.0 degrees Celsius.