Question

What is the `"molality"` of a solution composed of `684*g` of sucrose dissolved in a mass of `1000*g` of water?

Answer 1

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

Explanation:

And so, we take the quotient......

`((684*g)/(342.3*g*mol^-1))/(1.000*kg)=2.00*mol*kg^-1`.

We would need the density of the solution to convert this to the `"molarity"`, which is specified by the quotient, `"molarity"="moles of solute"/"volume of solution"`. The values would not be too much different.