What is the freezing point depression of an aquous solution of 10.0 g of glucose in 50.0 g H2O?

I need help

moles glucose = grams/molar mass.

Solve for moles glucose.

molality = moles/kg solvent.
Solve for molality

delta T = Kf*m
Solve for delta T, then convert to freezing point.

biggie smalls is a legend

Celtics in four

I'm sorry, how can I assist you with your query or concern?

Well, if you want a literal answer, the freezing point depression of a solution mainly depends on the concentration of solute particles. But let's not get all serious and scientific here!

Why did the ice cream get depressed?
Because it was feeling a little "low" due to the glucose! *wink* *wink*

All joking aside, we can calculate the freezing point depression using the equation:

ΔTf = Kf * m

Where ΔTf is the change in freezing point, Kf is the cryoscopic constant, and m is the molality of the solution.

Assuming the cryoscopic constant for water is 1.86 °C/molal, we substitute the values:

ΔTf = 1.86 °C/m * (mol solute / kg solvent)

We can find the molality (mol solute / kg solvent) by dividing the moles of glucose by the mass of water in kg.

Now, the molecular weight of glucose (C6H12O6) is approximately 180 g/mol. So, we first convert 10.0 g of glucose to moles:

10.0 g / 180 g/mol = 0.0556 mol

Next, we divide the mass of water by 1000 to convert it to kg:

50.0 g / 1000 g/kg = 0.0500 kg

Now we can calculate the molality:

molality = 0.0556 mol / 0.0500 kg = 1.11 mol/kg

Finally, we plug the values into the equation:

ΔTf = 1.86 °C/m * 1.11 mol/kg = 2.06 °C

So, the freezing point of the solution would be lowered by 2.06 °C. And that's the not-so-depressing answer!

To calculate the freezing point depression of a solution, you need to know the freezing point depression constant (K_f), the molality of the solute, and the molar mass of the solute.

1. Find the molality (m) of the solution:
Molality (m) is defined as the number of moles of solute per kilogram of solvent. In this case, the solute is glucose and the solvent is water.

First, calculate the number of moles of glucose:
Number of moles (n) = mass of glucose (m) / molar mass of glucose (M)

The molar mass of glucose (C6H12O6) is approximately 180.16 g/mol.
n = 10.0 g / 180.16 g/mol

Next, calculate the mass of the water (solvent):
Mass of water = mass of solution - mass of glucose = 50.0 g - 10.0 g

Finally, calculate the molality of the solution:
Molality (m) = n / (mass of water (in kg))

2. Find the freezing point depression (∆Tf):
The freezing point depression is given by the equation:
∆Tf = K_f * m

The value of the freezing point depression constant (K_f) depends on the solvent. For water, the K_f value is approximately 1.86 °C/m.

∆Tf = 1.86 °C/m * molality (m)

So, to find the freezing point depression of the aqueous solution of 10.0 g of glucose in 50.0 g of H2O, you need to calculate the molality of the solution and then use that value to find the freezing point depression (∆Tf).

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