# 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?

##### 1 Answer

#### Explanation:

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

#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

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

#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 **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**.