Question

What is the molality of a solution that contains 5 moles of solute in 100 kilograms of water?

Answer 1

Answer:
molality = (moles of solute) / (kilogram of solvent)

moles of solute = (75.2 g AgClO4) / (mole mass AgClO4)
moles of solute = (75.2 g AgClO4) / (207.3206 g/mol)
moles of solute = 0.3627232 mol

molality = (0.3627232 mol) / (0.0752 Kg)
molality = 4.8234468 m or 4.82 m rounded to three significant figures

Answer 2

`m=0.05m`

Step-by-step explanation:

Hello,

In this case, molality is defined as:

`m=(n_(solute))/(m_(solvent))`

Whereas the mass of the solvent is given in kilograms. In such a way, for 5 moles of solute and 100 kg of water, the molality turns out:

`m=(5mol)/(100kg)\\ m=0.05(mol)/(kg)`

Now it is important to notice that the molal units (m) equals mol/kg, thereby:

`m=0.05m`

Best regards.