Molality Problem

[matth87]

A solution is prepared containing 450 grams of NaI in 2000 grams of an unknown solvent. The mixture is stirred until the NaI has dissolved completely. What is the molality of this solution?

Can someone work this for me?

Thanks.

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[ktpinnock]

A solution is prepared containing 450 grams of NaI in 2000 grams of an unknown solvent. The mixture is stirred until the NaI has dissolved completely. What is the molality of this solution?

Can someone work this for me?

Thanks.

Molality = moles of solute/Kg of slovent

Moles of NaI = (450g) *(1mol/150g) = 3mols
Kg of Solvent = 2kg

Molality = 3mols/2kg or 1.5m

 

[matth87]

thanks

 

[inaccensa]

Molality = moles of solute/Kg of slovent

Moles of NaI = (450g) *(1mol/150g) = 3mols
Kg of Solvent = 2kg

Molality = 3mols/2kg or 1.5m

Since, NaI has dissolved, doesn't that imply that it dissociated? Then the molality should be 3 rt?

 

[crazybob]

Since, NaI has dissolved, doesn't that imply that it dissociated? Then the molality should be 3 rt?

1.5 m Na+ and 1.5 m I-

 

[wanderer]

!

 

[inaccensa]

1.5 m Na+ and 1.5 m I-

so it is 3m rt

 

[crazybob]

so it is 3m rt

why would it be 3 m? you have 450 grams of NaI (150 grams/mol) so 3 moles of NaI. NaI is the solute. you have 2000 grams of an unknown solvent. that's 2 kg of solvent.

molality is the ratio of moles of solute per kilogram of solvent. so 3 moles of NaI for 2 kilogram of solvent, which is 1.5 m.

if NaI dissociates completely, you have 3 moles of Na+ and 3 moles of I-. if you calculate molality with that, you get 1.5 m Na+ and 1.5 m I-. you cannot add the two together, because they would be for different solutes.