[webassignbuddy]

The boiling point of benzene (Mm = 78.0 g/mol; k_b = 2.53ºC/m; ρ = 0.877 g/mL) is 80.1ºC. What mass of vanillin (mm = 152.1 g/mol; non-electrolyte) is required as a solute to raise the boiling point of 44.6 mL of benzene to 81.2ºC?

Solvent (benzene)
Tº_b  = 80.1ºC
k_b  = 2.53ºC/m
ΔT_b = 81.2ºC
Mm = 78.9 g/mol
ρ = 0.877 g/mL
V = 44.6 mL

Solute (vanillin)
Mm = 152.1 g/mol
i = 1 (non-electrolyte)

ΔT_b = k_bm_c
ΔT_b = k_b(1 * m)
ΔT_b = k_b(1 * mol solute/mass kg solvent)
81.2 = (2.53)(mol solute/___ mass kg solvent)

Solve for mass kg solvent (benzene)
44.6 mL x 0.877 g/mL = 39.1142 g = 0.039 kg solvent, right?

Am I on the right track??

[Borek]

81.2 = (2.53)(mol solute/___ mass kg solvent)

Think it over.

[webassignbuddy]

81.2 = (2.53)(mol solute/___ mass kg solvent)

Think it over.

I did and I can't figure out how I got marked wrong up to here.
The final is tomorrow and I'm stressing out trying to figure out how to solve this one
I have no idea why they gave my Mm of benzene also!

[webassignbuddy]

81.2 = (2.53)(mol solute/___ mass kg solvent)

Think it over.

Oh wait nvm I figured it out.

1.1 = (2.53)(mol solute/0.039)

And then I just multiply mol solute by 152.1 g vanillin.