# What are the "molarities" and "molalities" of a 180xx10^3*g mass of glucose dissolved in 1*kg of water?

Jul 26, 2017

Unreasonably high.......

#### Explanation:

$\text{Molality"="Moles of solute"/"Kilograms of solvent}$

While we can find the molar quantity, i.e. $\text{moles of solute} = \frac{180 \times {10}^{3} \cdot g}{180.16 \cdot g \cdot m o {l}^{-} 1} = 1000 \cdot m o l$, the presumption that such a quantity would dissolve in approx. one litre of solution is unreasonable.

If the question specified a $180.16 \cdot g$ quantity, then I doubt that the volume of the aqueous solution would change much......

And so "molality"=((180.16*g)/(180.16*g*mol^-1))/(1*kg*"water")=1*mol*kg^-1. Would this solution be denser or less dense than pure water?