Question #01e41

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

In your case, a sucrose solution that has a molarity of #"1.6 M"# will contain #1.6# moles of sucrose, the solute, for every #"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 #"342.30 g mol"^(-1)#, which means that every mole of sucrose has a mass of #"342.30 g"#.

This means that your solution will contain

#0.56 color(red)(cancel(color(black)("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.