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?

"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

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 =)

"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

Sure, I can help you with that! To find the molality of a solution, you need to know the amount of solute (in moles) and the mass of the solvent (in kilograms).

In this case, you have 5.61 moles of KCl as the solute and 2.11 kg of water as the solvent. To calculate the molality, you can use the formula:

Molality (m) = moles of solute / mass of solvent (in kg)

First, convert the mass of water from kilograms to grams by multiplying by 1000:

2.11 kg * 1000 g/kg = 2110 g

Next, divide the moles of KCl by the mass of water in kg to get the molality:

Molality = 5.61 mol / 2.11 kg = 2.66 mol/kg

Therefore, the molality of the solution is 2.66 mol/kg.

M x L = M x L

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!