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?

Answers

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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