Question

How many moles are present in a 4.25 m aqueous solution prepared with 400. g of water?

Answer 1

Approx. `1.7*mol...`

Explanation:

`"Molality"="Moles of solute"/"Kilograms of solvent...."`

And so for a `4.25*mol*kg^-1`, there are `4.25*mol*kg^-1xx0.400*kg-=1.70*mol` with repsect to the solute....