What is the "molality"molality of a 0.030*g0.030⋅g mass of sugar dissolved in 300*mL300⋅mL of water?

1 Answer
anor277
May 3, 2017

"Molality"-="Moles of solute"/"Kilograms of solvent"Molality≡Moles of soluteKilograms of solvent

Explanation:

And so "Molality"=((0.03*g)/(180.16*g*mol^-1))/(0.3*kg)=??mol*kg^-1Molality=0.03⋅g180.16⋅g⋅mol−10.3⋅kg=??mol⋅kg−1.

At these concentrations the "molality"molality would be identical to the "molarity"molarity.

Note that 1*dm^31⋅dm3 is a fancy way of saying 1*L1⋅L, 1*dm^31⋅dm3 == "1 decimetre"1 decimetre, where the prefix "deci"-=10^-1deci≡10−1.

And thus 1*dm^3"-=(10^-1*m)^3=10^-3*m^3=1*L1⋅dm3≡(10−1⋅m)3=10−3⋅m3=1⋅L, because there are 1000*L*m^-31000⋅L⋅m−3. A "cubic metre"cubic metre is a huge volume.

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