Question

27.9 g of a solute raises the boiling point of benzene to 83.74 oC. The mass of the benzene solvent is 367 g and the solute is nonionic. Find the molecular weight of the solute.

Normal boiling point for benzene: 80.1oC; Kb = 2.53oC/m

Answer 1

Answer: To find the molecular weight of the solute, we can use the formula for calculating the elevation of boiling point (ΔTb) in a solution:

ΔTb = K_b * m

where:

ΔTb = elevation of boiling point (change in boiling point)

K_b = molal boiling point elevation constant (given as 2.53°C/m for benzene)

m = molality of the solution (moles of solute per kilogram of solvent)

First, let's find the molality (m) of the solution:

Molality (m) = moles of solute / mass of solvent (in kg)

Step 1: Calculate the moles of solute:

Given mass of the solute = 27.9 g

Moles of solute = mass of solute / molar mass of solute

Now, we need to find the change in boiling point (ΔTb). The normal boiling point of benzene is 80.1°C, and the boiling point after adding the solute is 83.74°C.

Step 2: Calculate the elevation of boiling point (ΔTb):

ΔTb = boiling point with solute - normal boiling point

ΔTb = 83.74°C - 80.1°C

ΔTb = 3.64°C

Step 3: Calculate the molality (m) of the solution:

m = moles of solute / mass of solvent (in kg)

Given mass of benzene solvent = 367 g

Convert the mass of benzene solvent to kilograms:

Mass of benzene solvent = 367 g / 1000 g/kg = 0.367 kg

Now, calculate the molality (m):

m = moles of solute / mass of solvent (in kg)

m = (27.9 g / molar mass of solute) / 0.367 kg

Step 4: Use the molality (m) to find the molecular weight of the solute:

m = 2.53°C/m (given boiling point constant for benzene)

Now, we can rearrange the formula to find the molar mass of the solute:

Molar mass of solute = (27.9 g / (0.367 kg * 2.53°C/m)

Molar mass of solute ≈ 73.2 g/mol

So, the molecular weight of the solute is approximately 73.2 g/mol.