GChem..help please...

[future_dentist]

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When 25.5 grams of nonvolatile nonelectrolyte are placed into 500 grams of water, the boiling point of solution at one atmosphere is 101.56 C. What is the molecular weight of the solute? (Kb of H2O = 0.52 C Kg/mole)

A) 102 g/mole
B) 51 g/mole
C) 34 g/mole
D) 17 g/mole
E) 8.5 g/mole

Ans: D

Thnxs...

 

[egots]

When 25.5 grams of nonvolatile nonelectrolyte are placed into 500 grams of water, the boiling point of solution at one atmosphere is 101.56 C. What is the molecular weight of the solute? (Kb of H2O = 0.52 C Kg/mole)

A) 102 g/mole
B) 51 g/mole
C) 34 g/mole
D) 17 g/mole
E) 8.5 g/mole

Ans: D

Thnxs...

1.56 = ((25.5/x)/.5)(.52)

1.56 is differrent water boils at 100... 1.56= (molality)(Kb)

molality is mol/kg if you have 25.5 grams given u divide that by X the molecular weight to find moles put that over the .5 kg...so use taht first equation solve for X

 

[future_dentist]

1.56 = ((25.5/x)/.5)(.52)

1.56 is differrent water boils at 100... 1.56= (molality)(Kb)

molality is mol/kg if you have 25.5 grams given u divide that by X the molecular weight to find moles put that over the .5 kg...so use taht first equation solve for X

Thank u...so much....
i really feel stupid it was such an easy question....i should have gotten it right 😡

U did a really good job with explaining.....i wish i had ppl.....from SDN teaching me Kaplan...because my teacher cant explain for jack.....

 

[egots]

no prob dont sweat it...when you dont see things for awhile you tend to forget it...just keep praticing the more you get wrong the more you learn, once u familiarize yourself with all the stuff itll come real easily...good luck