Question

How many grams of glucose must be added to 275 g of water in order to prepare percent by mass concentration of aqueous glucose solution 1.30% ?

Answer 1

`"3.62 g"`

Explanation:

The idea here is that the solution's percent concentration by mass tells you the number of grams of solute, which in your case is glucose, present for every `"100. g"` of the solution.

A solution that is `"1.30%` glucose by mass will contain `"1.30 g"` of glucose for every `"100. g"` of the solution.

So the question is actually asking about the number of grams of glucose that must be added to `"275 g"` of water in order to get `"1.30 g"` of glucose for every `"100. g"` of this solution.

If you take `x` `"g"` to be the mass of glucose needed to make this solution, you can say that after you add the glucose to the water, the total mass of the solution will be

`x quad "g" + "275 g" = (x + 275) quad "g"`

So you know that `x` `"g"` of glucose in `(x + 275)` `"g"` of the solution must be equivalent to `"1.30 g"` of glucose in `"100. g"` of the solution.

This means that you have

`(x color(red)(cancel(color(black)("g glucose"))))/((x + 275) color(red)(cancel(color(black)("g solution")))) = (1.30 color(red)(cancel(color(black)("g glucose"))))/(100. color(red)(cancel(color(black)("g solution"))))`

Rearrange and solve for `x` to get

`100. * x = 1.30 * x + 1.30 * 275`

`98.7 * x = 357.5 implies x = 357.5/98.7 = 3.6221`

Rounded to three sig figs, the number of sig figs you have for your values, the answer will be

`color(darkgreen)(ul(color(black)("mass of glucose = 3.62 g")))`