# 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% ?

Apr 18, 2018

$\text{3.62 g}$

#### 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 $\text{100. g}$ of the solution.

A solution that is "1.30% glucose by mass will contain $\text{1.30 g}$ of glucose for every $\text{100. g}$ of the solution.

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

If you take $x$ $\text{g}$ 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

$x \quad \text{g" + "275 g" = (x + 275) quad "g}$

So you know that $x$ $\text{g}$ of glucose in $\left(x + 275\right)$ $\text{g}$ of the solution must be equivalent to $\text{1.30 g}$ of glucose in $\text{100. g}$ of the solution.

This means that you have

$\left(x \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g glucose"))))/((x + 275) color(red)(cancel(color(black)("g solution")))) = (1.30 color(red)(cancel(color(black)("g glucose"))))/(100. color(red)(cancel(color(black)("g solution}}}}\right)$

Rearrange and solve for $x$ to get

$100. \cdot x = 1.30 \cdot x + 1.30 \cdot 275$

$98.7 \cdot x = 357.5 \implies x = \frac{357.5}{98.7} = 3.6221$

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

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{mass of glucose = 3.62 g}}}}$