Question

Question #356ae

Answer 1

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