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

Last updated date: 03rd Mar 2024
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Hint: The mass percent is utilized to communicate the concentration of a solution at which we know the mass of a solute and the mass of a solution:
${\text{Mass Percent}} = \dfrac{{{\text{Mass of Solute}}}}{{{\text{Mass of Solution}}}} \times 100{\text{% }}$
In science, the mass concentration ${\rho _i}$ (or ${\gamma _i}$) is characterized as the mass of a constituent ${m_i}$ partitioned by the volume of the combination $V$.
${\rho _i} = \dfrac{{{m_i}}}{V}$

The thought here is that the arrangement's percent concentration by mass reveals to you the quantity of grams of solute, which for your situation is glucose, present for each $100.{\text{ }}g$ of the arrangement.
A solution that is $1.30\%$ glucose by mass will contain $1.30{\text{ }}g$ of glucose for each $100.{\text{ }}g$ of the arrangement.
So, the inquiry is really posing about the quantity of grams of glucose that should be added to $275{\text{ }}g$ of water to get $1.30{\text{ }}g$ of glucose for each $100.{\text{ }}g$ of this arrangement.
On the off chance that you take $x{\text{ }}g$ to be the mass of glucose expected to make this arrangement, you can say that after you add the glucose to the water, the total mass of the arrangement or solution will be
$x\;\;{\text{g}} + {\text{275 g}} = \left( {x + 275} \right)\;\;{\text{g}}$
So, you realize that $x{\text{ }}g$ of glucose in $\left( {x + 275} \right){\text{ }}g$ of the arrangement should be comparable to $1.30{\text{ }}g$ of glucose in $100.{\text{ }}g$ of the arrangement.
This implies that you have
$\dfrac{{x\not{{{\text{g glucose}}}}}}{{\left( {x + 275} \right)\not{{{\text{g solution}}}}}} = \dfrac{{1.30\not{{{\text{g glucose}}}}}}{{100.\not{{{\text{g solution}}}}}}$
rearrange it and do calculation for $x$ to get
$100. \cdot x = 1.30 \cdot x + 1.30 \cdot 275$
$98.7 \cdot x = 357.5 \Rightarrow x = \dfrac{{357.5}}{{98.7}} = 3.6221$
Adjusted to three sig figs, the quantity of sig figs you have for your qualities, the appropriate answer will be
$\left[ {{\text{mass of glucose = 3}}{\text{.62 g}}} \right]$

Note: Solution concentration changes when solution volume changes if there ought to emerge an event of molarity similarly as in case of focus by mass considering the way that the mass of a substance will change its volume, in this manner the focus changes.
${\text{Mass Percent}} = \dfrac{{{\text{Mass of Solute}}}}{{{\text{Mass of Solution}}}} \times 100{\text{% }}$ (For mass percentage)
${\text{Volume Percent}} = \dfrac{{{\text{Volume of Solute}}}}{{{\text{Volume of Solution}}}} \times 100{\text{% }}$(For volume percentage)