# Question 4f37f

Feb 27, 2017

${\text{0.056 mol L}}^{- 1}$

#### Explanation:

You know that you're dealing with a $\text{1% w/v}$ glucose solution, which essentially means that every $\text{100 mL}$ of solution will contain $\text{1 g}$ of glucose.

Now, a solution's molarity is determined by taking the number of moles of solute present in $\text{1 L}$ of solution.

To make the calculations easier, select a $\text{1.0-L}$ sample of this glucose solution. The solution's mass by volume percent concentration tells you that

1.0 color(red)(cancel(color(black)("L solution"))) * (10^3"mL")/(1color(red)(cancel(color(black)("L")))) = 10^3"mL"

of solution will contain

10^3 color(red)(cancel(color(black)("mL solution"))) * "1 g glucose"/(100color(red)(cancel(color(black)("mL solution")))) = "10 g glucose"

Now all you have to do is convert this to moles of glucose by using the compound's molar mass

10 color(red)(cancel(color(black)("g"))) * "1 mole glucose"/(180color(red)(cancel(color(black)("g")))) = "0.0556 moles glucose"#

Since this represents the number of moles of glucose present in $\text{1.0 L}$ of solution, you can say that the solution has a molarity of

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molarity = 0.056 mol L}}^{- 1}}}}$

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the percent concentration.