Chemistry: Boiling Pt. Elevation and Molality related question

[pandalove89]

When 25.5 grams of a nonvolatile nonelectrolyte are placed into 500 grams of water, the boiling point of the solution at one atmosphere is 101.56 Celsius. What is the molecular weight of the solute? (Kb of H2O = 0.52 K*kg/mol)


The answer is 17 g/mol but I have no idea how to do this...

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[newbie1990]

Solve for molality in the boiling point elevation equation:

(1.56) = (.52)X


Since molality = moles of solute/kg of solvent, you can set up another equation solving for the molecular weight (because moles = g/molecular weight).

X = [(25.5 grams)/MW]/(.5 kg)

I think this is how you'd set it up... does this make sense? haha it's kinda confusing to type it all out on the computer...

 

[Mujtaba]

When 25.5 grams of a nonvolatile nonelectrolyte are placed into 500 grams of water, the boiling point of the solution at one atmosphere is 101.56 Celsius. What is the molecular weight of the solute? (Kb of H2O = 0.52 K*kg/mol)


The answer is 17 g/mol but I have no idea how to do this...

Use the formula ΔTb = iKbm. This gives you the change in temperature from the boiling point. i is the number of particles dissolved in the solvent. So for example, i=2 for NaCl from Na+ and Cl+ ions. In this example, it tells us a nonelectrolyte is placed in the solution (meaning it doesn't break apart), so i=1. They give you Kb, and m represents molality ( mol solute/ kg of solvent).

So you would set up the problem like this:
25.5 g / (x g/mol * .5 kg) * .52 = 1.56
Solving for x gets you 17.

 

[dentcal]

sry u lost me. where did you get the 1.56 from?

 

[LetsGo2DSchool]

sry u lost me. where did you get the 1.56 from?

1.56 is change in temp between boiling point elevation temp. (101.56) and boiling point of water (100).

 

[dentcal]

haha wow for got about the delta!