genchem question...

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can any1 help me thought this problem? its from the KAPLAN DAT CD # 61.

-when 25.5 grams of a nonvolatile nonelectrolyte are placed in 500g of H20, the boiling point of the solution at 1 ATM is 101.56C. What is the molecular weight of the solute?

a. 102 g/mol
b. 51 g/mol
c. 34 g/mol
d. 17 g/mol --> answer
e. 8.5 g/mol

 

[askadds]

the Kb of H20 = .52 C*kg / mol

can any1 help me thought this problem? its from the KAPLAN DAT CD # 61.

-when 25.5 grams of a nonvolatile nonelectrolyte are placed in 500g of H20, the boiling point of the solution at 1 ATM is 101.56C. What is the molecular weight of the solute?

a. 102 g/mol
b. 51 g/mol
c. 34 g/mol
d. 17 g/mol --> answer
e. 8.5 g/mol

 

[DrWanahbe]

OK, here's what I did:

The normal boiling point of a water is 100 degrees. You are told the bp is 101.56. First, find the difference, which is 1.56. This is your delta T.

Then, rearrage the equation delta T = kb*m(solute) to solve for m(solute). So, 1.56/kb, which you are given as .51, gives you 3 moles/kg (molality)

To find moles, you use the molality equation moles solute/kg of solution, so
3*.500kg gives you 1.5 moles.

Molecular weight is g/moles. You know you have 25.5 grams and 1.5 moles. Divide 25.5 by 1.5 and you get 17 g/mol.

Hope that helps!

 

[askadds]

thanks for the help!