Question

Question #5b4e9

Answer 1

`15.3%`

Explanation:

A solution's molality tells you how many moles of solute you get for every `"1 kg"` of solvent.

In this regard, a `"1 m"` solution will contain `1` mole of glucose, your solute, for every `"1 kg"` of water, your solvent. Since you know that you have exactly `"1 kg"` of water, you can say that this solution contains exactly `1` mole of glucose.

Now, in order to find the solution's percent concentration by mass, `"% m/m"`, you must figure out the number of grams of solute present for every `"100 g"` of solution.

To convert the number of moles of glucose to grams, use the compound's molar mass

`1 color(red)(cancel(color(black)("mole glucose"))) * "180.156 g"/(1color(red)(cancel(color(black)("mole glucose")))) = "180.156 g"`

So, you know that you have `"180.156 g"` of glucose in exactly

`1 color(red)(cancel(color(black)("kg"))) * (10^3color(white)(.)"g")/(1color(red)(cancel(color(black)("kg")))) = "1000 g"`

of water, which means that the total mass of the solution will be

`m_"total" = "180.156 g" + "1000 g" = "1180.156 g"`

You can now use the known composition of the solution to determine how many grams of glucose would be present in `"100 g"` of solution

`100 color(red)(cancel(color(black)("g solution"))) * "180.156 g glucose"/(1180.156color(red)(cancel(color(black)("g solution")))) = "15.3 g glucose"`

Since this is how many grams of glucose you have for every `"100 g"` of solution, you can say that the solution's percent concentration by mass is equal to

`color(darkgreen)(ul(color(black)("% m/m" = 15.3%)))`

I'll leave the answer rounded to three sig figs, but keep in mind that you only have one significant figure for the molality of the solution and for the mass of water.