Question #356ae

1 Answer
Jul 3, 2017

Answer:

Here's what I got.

Explanation:

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 #"1 L"# of #"1.4-M"# solution.

The thing to remember about molarity is that it represents the number of moles of solute present in #"1 L"# of solution. In your case, a #"1.4-M"# glucose solution will contain #1.4# moles of glucose for every #"1 L"# of solution.

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 #"1.4-M"# glucose solution will contain #"250 g"# of glucose, the equivalent of #1.4# moles of glucose, for every #"1 L"# of solution.

You can use this information to figure out the mass of glucose that would be present in, for example, #"0.5 L"# of #"1.4-M"# solution.

#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