# Question #5b4e9

##### 1 Answer

#### Explanation:

A solution's **molality** tells you how many *moles* of solute you get **for every** **of solvent**.

In this regard, a **mole** of glucose, your solute, for every **mole** of glucose.

Now, in order to find the solution's **percent concentration by mass**, *grams* of solute present **for every** **of solution**.

To convert the number of moles of glucose to grams, use the compound's **molar mass**

#1 color(red)(cancel(color(black)("mole glucose"))) * "180.156 g"/(1color(red)(cancel(color(black)("mole glucose")))) = "180.156 g"#

So, you know that you have

#1 color(red)(cancel(color(black)("kg"))) * (10^3color(white)(.)"g")/(1color(red)(cancel(color(black)("kg")))) = "1000 g"#

of water, which means that the **total mass** of the solution will be

#m_"total" = "180.156 g" + "1000 g" = "1180.156 g"#

You can now use the known composition of the solution to determine how many grams of glucose would be present in

#100 color(red)(cancel(color(black)("g solution"))) * "180.156 g glucose"/(1180.156color(red)(cancel(color(black)("g solution")))) = "15.3 g glucose"#

Since this is how many grams of glucose you have **for every** **of solution**, you can say that the solution's percent concentration by mass is equal to

#color(darkgreen)(ul(color(black)("% m/m" = 15.3%)))#

I'll leave the answer rounded to three **sig figs**, but keep in mind that you only have one significant figure for the molality of the solution and for the mass of water.