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.