Question #4f37f

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

Now, a solution's molarity is determined by taking the number of moles of solute present in #"1 L"# of solution.

To make the calculations easier, select a #"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 #"1.0 L"# of solution, you can say that the solution has a molarity of

#color(darkgreen)(ul(color(black)("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.