Question

Question #01e41

Answer 1

`"190 g"`

Explanation:

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.