Asked by Alan

What is the molality of a solution made from 5.61 moles of KCl and 2.11 kg of water?

I'm very lost on this one, could someone please help me?
Answered by DrBob222
Just follow the definition. A 1 molal solution is defined as 1 mole of material dissolved in 1 kg solvent; therefore, molality = # moles/kg solvent.
You have moles given and you have kg solvent. Voila!
Answered by Alan
How much concentrated 18 M sulfuric acid is need to prepare 250 ml of a 6.0 M solution?

This is the last one guys, its got me stumped again =)
Answered by DrBob222
M x L = M x L
Answered by Nick
"5.61 moles of KCl and 2.11 kg of water?" Well, water has a density of 1g/ml. So, 2.11 kg, is 2.11 liters. Therefore you have 5.61 moles in 2.11 liters. Molar is moles per liter, therefore, 5.61 / 2.11 = 2.66 M. For help on moles and molar have a look at cal.culate.it/moles
Answered by Nick
"18 M sulfuric acid is need to prepare 250 ml of a 6.0 M solution" - this is an M1V1 = M2V2 type problem. For help have a look at cal.culate.it/dilution. Meanwhile, using chemCal on the iPhone (cal.culate.it/c) the answer is: 83.3 ml
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