Question

What is the molality of a solution that contains `"3 g"` of glucose dissolved in `"30 g"` of water?

Answer 1

`"0.6 mol kg"^(-1)`

Explanation:

Molality is all about moles of solute dissolved per kilogram of solvent. In other words, all you have to do in order to calculate a solution's molality is to find out how many moles of solute you have per kilogram of solvent.

Use the molar mass of glucose, `"C"_6"H"_12"O"_6`, to determine how many moles are present in your sample

`3 color(red)(cancel(color(black)("g"))) * ("1 mole C"_6"H"_12"O"_6)/(180.16color(red)(cancel(color(black)("g")))) = "0.01665 moles C"_6"H"_12"O"_6`

Now, you know that this many moles of glucose, your solute, are dissolved in `"30 g"` of water, which is your solvent.

Your goal now is to "scale up" this solution to `"1 kg"` of water by keeping the given proportion between the number of moles of solute and the mass of solvent.

Convert the mass of water from grams to kilograms

`30 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "0.030 kg"`

So, if `"0.030 kg"` of water contain `0.01665` *moles8 of glucose, it follows that `"1 kg"` of water will contain

`1 color(red)(cancel(color(black)("kg water"))) * ("0.01665 moles C"_6"H"_12"O"_6)/(0.030color(red)(cancel(color(black)("kg water")))) = "0.555 moles C"_6"H"_12"O"_6`

Therefore, you can say that the molality of the solution is

`"molality" = color(green)(|bar(ul(color(white)(a/a)color(black)("0.6 mol kg"^(-1) = "0.6 molal")color(white)(a/a)|)))`

The answer must be rounded to one sig fig, the number of sig figs given to you for the mass of glucose and the mass of water.