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What mass in grams of CH₃OH (MM = 32.04 g/mol) would need to be added to 380.1 g of water in order to lower the freezing point of water by 12.0 °C? (Kf for water is 1.86 °C/m)

User Kdubs
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Final answer:

To calculate the mass of CH₃OH needed to lower the freezing point of water by 12.0 °C, you can use the formula: ΔT = Kf x m. Rearrange the formula and solve for m to find the molarity of the solution. Convert the molarity to moles and then to grams using the molar mass of CH₃OH.

Step-by-step explanation:

Answer:


To calculate the mass of CH₃OH needed to lower the freezing point of water, we can use the formula:

ΔT = Kf x m
In this case, we know that ΔT is -12.0 °C, Kf for water is 1.86 °C/m, and we want to find the molarity of the solution. Rearranging the formula, we can solve for m:

m = ΔT / Kf = -12.0 °C / 1.86 °C/m = -6.45 m

Since molarity (m) is defined as moles of solute per liter of solution, we need to convert the molarity to moles:

moles of solute = m x liters of solution

First, we convert the mass of water to liters using its density of 1 g/mL:

380.1 g x (1 mL/1 g) x (1 L/1000 mL) = 0.3801 L

Now, we can calculate the moles of solute:

moles of solute = -6.45 m x 0.3801 L = -2.454 moles

Finally, we can convert moles of solute to grams using the molar mass of CH₃OH:

mass of solute = moles x molar mass = -2.454 moles x 32.04 g/mol = -78.52 g

Since we can't have a negative mass, it seems there may be an error in the calculation.

Double-check the given values and calculations to ensure accuracy.

If the given values are correct, then the calculated mass should be positive instead of negative.

User Taugenichts
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