What is the molarity of a sucrose solution that contains 10.0 g of `C_12H_22O_11` (342.34 g/mol) dissolved in 100.0 mL of solution?

Your goal when trying to find a solution's molarity is to determine how many moles of solute you have in one liter of solution.

Notice that your solution has a volume of `"100.0 mL"`. Since

`color(purple)(|bar(ul(color(white)(a/a)color(black)("1 L" = 10^3"mL")color(white)(a/a)|)))`

you can say that your solution has a volume that is equivalent to `1/10"th"` of `"1 L"`. Therefore, the number of moles that will be present in your sample will represent `1/10"th"` of the number of moles present in `"1 L"` of this solution.

So, use sucrose's molar mass to find the number of moles present in your sample

`10.0 color(red)(cancel(color(black)("g"))) * "1 mole sucrose"/(342.34color(red)(cancel(color(black)("g")))) = "0.02921 moles sucrose"`

So, if this is how many moles you have in `"100.0 mL"` of this solution, it follows that `"1 L"` will contain

`1 color(red)(cancel(color(black)("L solution"))) * (10^3color(red)(cancel(color(black)("mL solution"))))/(1color(red)(cancel(color(black)("L solution")))) * "0.02921 moles"/(100color(red)(cancel(color(black)("mL solution")))) = "0.2921 moles"`

You get `0.2921` moles of sucrose, you solute, per liter of solution, which means that the solution's molarity will be

`"molarity" = c = color(green)(|bar(ul(color(white)(a/a)"0.292 mol L"^(-1)color(white)(a/a)|)))`

The answer is rounded to three sig figs.