# 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?

May 8, 2016

${\text{0.292 mol L}}^{- 1}$

#### 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 $\text{100.0 mL}$. Since

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 L" = 10^3"mL}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

you can say that your solution has a volume that is equivalent to $\frac{1}{10} \text{th}$ of $\text{1 L}$. Therefore, the number of moles that will be present in your sample will represent $\frac{1}{10} \text{th}$ of the number of moles present in $\text{1 L}$ 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 $\text{100.0 mL}$ of this solution, it follows that $\text{1 L}$ 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.