# Question 356ae

Jul 3, 2017

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 $\text{1 L}$ of $\text{1.4-M}$ solution.

The thing to remember about molarity is that it represents the number of moles of solute present in $\text{1 L}$ of solution. In your case, a $\text{1.4-M}$ glucose solution will contain $1.4$ moles of glucose for every $\text{1 L}$ 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 $\text{1.4-M}$ glucose solution will contain $\text{250 g}$ of glucose, the equivalent of $1.4$ moles of glucose, for every $\text{1 L}$ of solution.

You can use this information to figure out the mass of glucose that would be present in, for example, $\text{0.5 L}$ of $\text{1.4-M}$ 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