# Question #0c3b7

##### 1 Answer

Here's what I got.

#### Explanation:

The idea here is that you need to use the known composition of the glucose solution to determine how many grams of glucose are present in

So, you know that

#"1 L" = 10^3# #"mL"#

and that *liters* to *milliliters* and set up the known composition as a **conversion factor**.

#3.1 color(red)(cancel(color(black)("L solution"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "5.9 g glucose"/(100. color(red)(cancel(color(black)("mL solution")))) = "183 g glucose"#

Now, the problem should provide you with the number of *calories* or *kilocalories* released **per gram** of glucose.

Assuming that

#183 color(red)(cancel(color(black)("g glucose"))) * "3.811 kcal"/(1color(red)(cancel(color(black)("g glucose")))) = color(darkgreen)(ul(color(black)(7.0 * 10^2color(white)(.)"kcal")))#

The answer is rounded to two **sig figs**.