# Question 01e41

Feb 11, 2018

$\text{190 g}$

#### Explanation:

The molarity of a solution is simply a measure of the number of moles of solute present in exactly $\text{1 L" = 10^3 quad "mL}$ of the solution.

In your case, a sucrose solution that has a molarity of $\text{1.6 M}$ will contain $1.6$ moles of sucrose, the solute, for every $\text{1 L" = 10^3 quad "mL}$ of the solution.

So you can say that you have

"1.6 M sucrose " = " ""1.6 moles sucrose"/(10^3 quad "mL solution")

To find the number of moles present in your sample, you can use the molarity of the solution as a conversion factor.

350 color(red)(cancel(color(black)("mL solution"))) * "1.6 moles sucrose"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.56 moles sucrose"#

Finally, to convert the number of moles of sucrose to grams, you need to use the molar mass of the compound. Sucrose has a molar mass of ${\text{342.30 g mol}}^{- 1}$, which means that every mole of sucrose has a mass of $\text{342.30 g}$.

This means that your solution will contain

$0.56 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles sucrose"))) * "342.30 g"/(1color(red)(cancel(color(black)("mole sucrose")))) = color(darkgreen)(ul(color(black)("190 g}}}}$

The answer is rounded to two sig figs.