Question #b7992

1 Answer
Oct 20, 2017


#"250 g"#


For starters, you know that glucose is a nonelectrolyte, which implies that it does not dissociate when dissolved in water.

This means that the number of osmoles of glucose produced in solution will be equal to the number of moles of glucose dissolved to make the solution.

Consequently, you can say that osmolarity of glucose in this solution will be equivalent to its molarity.

So, you know that your solution must have a molarity of #"280 mmoles L"^(-1)# and that it must have a volume of #"5 L"#. The first thing to do here is to figure out exactly how many moles of glucose must be present in this solution--keep in mind that you have #"1 mole" = 10^3# #"mmoles"#.

#5 color(red)(cancel(color(black)("L solution"))) * (280 color(red)(cancel(color(black)("mmoles glucose"))))/(1color(red)(cancel(color(black)("L solution")))) * "1 mole"/(10^3color(red)(cancel(color(black)("mmoles")))) = "1.4 moles glucose"#

Now all you have to do is to use the molar mass of glucose to convert the number of moles to grams

#1.4 color(red)(cancel(color(black)("moles glucose"))) * "180.16 g"/(1color(red)(cancel(color(black)("mole glucose")))) = color(darkgreen)(ul(color(black)("250 g glucose")))#

I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the volume of the solution.