Here's what I got.
Since you didn't provide an actual volume for your target solution, let's take the easy route and calculate the mass of glucose needed to make
The thing to remember about molarity is that it represents the number of moles of solute present in
Now, you can use the molar mass of glucose to calculate the number of grams that would contain that many moles of glucose.
#1.4 color(red)(cancel(color(black)("moles glucose"))) * "180.156 g"/(1color(red)(cancel(color(black)("mole glucose")))) = "250 g" ->#rounded to two sig figs
You can thus say that a
You can use this information to figure out the mass of glucose that would be present in, for example,
#0.5 color(red)(cancel(color(black)("L solution"))) * "250 g glucose"/(1color(red)(cancel(color(black)("L solution")))) = "130 g glucose " ->#two sig figs