A solution of toluene in 801 g of cyclohexane has a boiling point of 90.3 °C. How many moles of toluene are in the solution? (For cyclohexane Kb = 2.92 °C/m, Tb = 80.9 °C)

The moles of toluene in 801 grams of the solution have been 2.578 mol.

For heating, a solution, the change in temperature can be given by:

T = i m

T = Change in temperature = Final temperature - Initial temperature

T = 90.3 - 80.9

T = 9.4

i = von't Hoff factor = 1

= 2.92/m

m = molality

9.4 = 1 2.92 m

m = 3.21 m

The molality of the solution of toluene has been 3.21 mol.

The molality can be given as moles per kg solvent.

The mass of solvent = 801 g. = 0.801 kg

molality =

3.21 =

Moles of toluene = 3.21 0.801

Moles of toluene = 2.578 mol.

The moles of toluene in 801 grams of the solution have been 2.578 mol.

For more information about moles in a solution, refer to the link:

link

We have 2.58 moles toluene in solution

Explanation:

Step 1: Data given

Mass of toluene = 801 grams

Boiling point of the solution = 90.3 °C

Boiling point elevation constant = 2.92 °C/m

Boiling point temperature = 80.9 °C

Step 2: Calculate molality

ΔT = i*Kb*m

⇒with ΔT = the boiling point elevation = 90.3-80.9 = 9.4 °C

⇒with i = the van't Hoff factor = 1

⇒with Kb = the Boiling point elevation constant = 2.92 °C/m

⇒with m = the molality

m = 9.4 °C / 2.92

m = 3.22 molal = moles / kg

Step 3: Calculate moles toluene

Molality = moles tolune / mass cyclohexane

3.22 molal = moles toluene / 0.801 kg

Moles toluene = 2.58 moles

We have 2.58 moles toluene in solution

i would say b

explanation:

but i think i would need more information than that

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.