calculate the number of grams of glycerol, C3H5(OH)3 (MW=92.1 g/mol), that must be dissolved in 520 grams of water to raise the boiling point to 102.00 degrees Celsius.

this whole chapter really lost me.

delta T = Kb*molality
delta T = 2 degrees.
Kb = boiling point elevation constant.
molality = calculate
Then convert molality to grams glycerol.

thanks, but what is MW in the question? is that just etra info to throw me off?

No, it isn't an extra and you must use it. After calculating molality, then you must determine the grams glycerol in 520 g H2O to make that molarity.
kg soln x molality x molar mass = grams

Right

In this question, MW stands for molar mass, which is the mass of one mole of a substance. It is not just extra information to throw you off, but rather a crucial value that you need to use in your calculations.

To calculate the molality (m) of the glycerol solution, you can use the formula:

molality = ΔT / Kb

where ΔT represents the change in boiling point (2 degrees Celsius) and Kb is the boiling point elevation constant.

To find the number of grams of glycerol (C3H5(OH)3) that must be dissolved in 520 grams of water, you need to convert molality (m) to grams using the formula:

grams = kg soln x molality x molar mass

Now, let's calculate step by step:

Step 1: Calculate the molality (m) using the given values of ΔT and Kb:
molality = 2°C / Kb

Since the boiling point elevation constant (Kb) is not provided in the question, you will need to refer to a table or look it up. For water, the Kb value is approximately 0.512°C/m.

molality = 2°C / 0.512°C/m ≈ 3.91 m

Step 2: Convert molality (m) to grams using the given mass of water (520 grams):
grams = (520 g / 1000 g/kg) x 3.91 m x 92.1 g/mol

grams = 0.520 kg x 3.91 m x 92.1 g/mol

grams ≈ 184.2 grams

Therefore, approximately 184.2 grams of glycerol (C3H5(OH)3) must be dissolved in 520 grams of water in order to raise the boiling point to 102.00 degrees Celsius.

MW in the question refers to the molar mass of glycerol (C3H5(OH)3), which is provided as 92.1 g/mol. This is not extra information, but rather a necessary piece of data for the calculation.

To calculate the number of grams of glycerol that must be dissolved in 520 grams of water, the following steps can be followed:

1. Calculate the boiling point elevation (delta T):
delta T = Kb * molality

In this case, the boiling point elevation constant (Kb) is not provided, but it can be looked up for water. Typically, for water, Kb is around 0.512 °C/m. However, it may vary depending on the context of your problem.

Given that the boiling point elevation is 2 degrees (delta T = 2 °C), you can rearrange the equation to solve for molality:
molality = delta T / Kb

2. Once you have the molality, you can convert it to grams of glycerol:
kg soln * molality * molar mass = grams

First, convert the mass of water from grams to kilograms:
520 grams of water = 520 g / 1000 = 0.52 kg of water

Then, plug in the values:
grams of glycerol = 0.52 kg * molality * molar mass

Make sure that your molality is in moles of solute per kilogram of solvent. If necessary, convert it to the proper units before using it in the equation.

By following these steps, you can calculate the number of grams of glycerol that must be dissolved in 520 grams of water to raise the boiling point to 102.00 degrees Celsius.