Question

Question #4f37f

Answer 1

`"0.056 mol L"^(-1)`

Explanation:

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.