How many grams of methanol are needed to raise the boiling point of 250 mL of water by 2.7C. The boiling point elevation constant for water = 0.512 C/m. Assume water to have a density of 1g/mL)
delta T = Kb*molality
2.7 = 0.512*m
solve for m
molality = moles/kg solvent
You have molality, kg = 0.250, solve for moles.
moles = g/molar mass. solve for g.
I don't think this is a good problem because the depression of freezing point and elevation of boiling point are based on NON-VOLATILE solutes and methanol is hardly non-volatile.
To determine the number of grams of methanol needed, we need to use the boiling point elevation equation:
ΔTb = K * m * i
Where:
ΔTb = boiling point elevation
K = boiling point elevation constant
m = molality of the solute
i = van 't Hoff factor (depends on the number of particles produced when the solute dissolves)
First, let's calculate the molality (m) of the solute (methanol):
Molality (m) = moles of solute / mass of solvent (in kg)
We are given:
Volume of water (solvent) = 250 mL = 0.25 L
Density of water = 1 g/mL
Mass of water = 0.25 L * 1 g/mL = 0.25 kg
Now, we need to calculate the moles of methanol. To do this, we'll need to know the molar mass of methanol (CH3OH), which is approximately 32.04 g/mol.
Next, we calculate the moles of methanol:
Moles of methanol = (grams of methanol) / (molar mass of methanol)
Then, we calculate the molality:
Molality (m) = (moles of methanol) / (mass of water in kg)
Now, we substitute the values into the boiling point elevation equation:
ΔTb = (K) * (m) * (i)
We are given:
ΔTb = 2.7°C
K = 0.512°C/m
i = 1 (since methanol does not dissociate in water)
Rearranging the equation, we can solve for the grams of methanol:
(grams of methanol) = (ΔTb) * (mass of water) / [(K) * (m) * (i)]
Plug in the given values to calculate the grams of methanol needed.