Question

The freezing point \langleT_{f}\rangle for t-butanol is 25.50 C and K_{f} is 9.1 deg * C / m Usually t-butanol absorbs water on exposure to the air. If the freezing point of a 14.8-g sample of t-butanol is measured as 24.59 ^ C how many grams of water are present in the sample?

Answer 1

0.359 grams

Step-by-step explanation:

We can use the freezing point depression formula to calculate the amount of water present in the sample of t-butanol:

ΔTf = Kf * molality

where ΔTf is the change in freezing point, Kf is the freezing point depression constant for t-butanol, and molality is the concentration of the solute in moles per kilogram of solvent.

First, we need to calculate the molality of the t-butanol solution:

molality = moles of solute / mass of solvent in kg

Since t-butanol is the solvent in this case, we will assume that the mass of t-butanol is equal to the mass of the sample:

mass of t-butanol = 14.8 g

Next, we need to calculate the moles of t-butanol using its molar mass:

molar mass of t-butanol = 74.12 g/mol

moles of t-butanol = mass of t-butanol / molar mass of t-butanol

moles of t-butanol = 14.8 g / 74.12 g/mol

moles of t-butanol = 0.1995 mol

Now we can use the freezing point depression formula to calculate the molality of the solution:

ΔTf = Kf * molality

ΔTf = 25.50 °C - 24.59 °C = 0.91 °C

molality = ΔTf / Kf

molality = 0.91 °C / 9.1 °C/m

molality = 0.1 mol/kg

Finally, we can calculate the mass of water present in the sample using the molality and the mass of t-butanol:

molality = moles of solute / mass of solvent in kg

0.1 mol/kg = moles of water / (mass of water / 1000 g)

moles of water = 0.1 mol/kg * (14.8 g / 74.12 g/mol)

moles of water = 0.0199 mol

mass of water = moles of water * molar mass of water

mass of water = 0.0199 mol * 18.02 g/mol

mass of water = 0.359 g

Therefore, there are approximately 0.359 grams of water present in the sample.

ALLEN