Question

51.7 g of a nonionic solute in 420 g water raises the boiling point to 101.3°C. Find the molecular weight of the solute.

Answer 1

Do you know the formula used for boiling point elevation? Also the constant K_b used for water.

Answer 2

nope they gave me this chart though

Answer 3

The formula looks like this:
\( \Delta T = K_b m \) Delta T = change in Temperature (from 100 C), K_b is a constant, and m is the molality of the solution.

Answer 4

ok but how do i use it?

Answer 5

You know that molality = moles of solvent / mass of solute?

Answer 6

nope i have done chem since freshman year I'm making up second semester online

Answer 7

OK, these would be good review terms: molality = moles solute / mass solvent. molarity = moles solute / volume solvent.
And we can also write that moles = mass (g) / molecular weight (g/mol) which is just the conversion of moles.

Answer 8

What we are looking to do though is set up:
\( \Delta T = K_b m \) \(m = \dfrac{mass}{molecular \ weight} = \dfrac{\bar{m}}{W} \)
\( \Delta T = K_b \dfrac{\bar{m}}{W} \)
Calling m with a bar the mass and W the molecular weight. We then plug in our information.
Delta T = 101.3 C - 100.0 C (change in temperature)
K_b = 0.512 C m^-1
\(\bar{m}\) = 51.7 g
and we want to find W.

Answer 9

can u slow down to fast

Answer 10

Sure. Which part should I have gone over more slowly?

Answer 11

the whole delta crap its confusing

Answer 12

Delta just means a change.
\( \Delta T = T_2 - T_1 \)
So in our case, we are looking at how temperature changes from normal boiling point at 100.0 degrees C. It is raised to 101.3 degrees C. The addition of the solute raises it from 100.0 to 101.3, so Delta T = 101.3 - 100.0 = 1.3 degrees C.

Sorry just got disconnected for a bit.