# Question 6d84d

Feb 13, 2016

$\text{14.5%}$

#### Explanation:

As you know, percent concentration by mass is defined as the mass of the solute, which in your case is sugar, divided by the total mass of the solution, and multiplied by $100$.

$\textcolor{b l u e}{\text{% w/w" = "mass of solute"/"mass of solution} \times 100}$

The problem provides you with the mass of the solution, which is said to be equal to

${m}_{\text{sol" = "5.2752 g}}$

You will also need to use the density of water, which at room temperature is equal to

${\rho}_{w} = {\text{0.9982 g mL}}^{- 1}$

http://antoine.frostburg.edu/chem/senese/javascript/water-density.html

and the density of solid sugar, or sucrose, which is listed as

${\rho}_{s} = {\text{1.587 g mL}}^{- 1}$

Now, your goal here will be to find the mass of solute, ${m}_{s}$.

As you know, density is defined as mass per unit of volume, which means that you can express volume using density and mass

$\textcolor{b l u e}{\rho - \frac{m}{V} \implies V = \frac{m}{\rho}}$

Now, the total volume of the solution will be equal to the volume of sugar, ${V}_{s}$, plus the volume of water, ${V}_{w}$

${V}_{\text{sol}} = {V}_{s} + {V}_{w}$

This is equivalent to

V_"sol" = m_s/(rho_s) + m_w/(rho_w)" " " "color(red)("(*)")

Here ${m}_{s}$ and ${m}_{w}$ represent the mass of sugar and the mass of water, respectively.

You also know that the total mass of solution, ${m}_{\text{sol}}$, is equal to

${m}_{\text{sol}} = {m}_{s} + {m}_{w}$

This means that the mass of water can be written as

${m}_{w} = {m}_{\text{sol}} - {m}_{s}$

Plug this into equation $\textcolor{red}{\text{(*)}}$ to get

${V}_{\text{sol" = m_s/(rho_s) + m_"sol}} / \left({\rho}_{w}\right) - {m}_{s} / \left({\rho}_{w}\right)$

Rearrange to isolate ${m}_{s}$ on one side of the equation

${m}_{s} \cdot \left(\frac{1}{\rho} _ s - \frac{1}{\rho} _ w\right) = {V}_{\text{sol" - m_"sol}} / {\rho}_{w}$

${m}_{s} \cdot \left(\frac{{\rho}_{w} - {\rho}_{s}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{\rho}_{w}}}} \cdot {\rho}_{s}}\right) = \frac{{V}_{\text{sol" * rho_w - m_"sol}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{\rho}_{w}}}}}$

Finally, you will get

${m}_{s} = \left({V}_{\text{sol" * rho_w - m_"sol}}\right) \cdot {\rho}_{s} / \left({\rho}_{w} - {\rho}_{s}\right)$

Plug in your values to get the value of ${m}_{s}$

${m}_{s} = \left(5.00 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{mL"))) * "0.9982 g" color(red)(cancel(color(black)("mL"^(-1)))) - "5.2752 g") * (0.1.587 color(red)(cancel(color(black)("g mL"^(-1)))))/(0.9982color(red)(cancel(color(black)("g mL"^(-1)))) - 1.587color(red)(cancel(color(black)("g mL}}^{- 1}}}}\right)$

${m}_{s} = \text{0.766 g}$

This means that the solution's percent concentration by mass is

"% w/w" = (0.766 color(red)(cancel(color(black)("g"))))/(5.2752color(red)(cancel(color(black)("g")))) xx 100 = color(green)("14.5%") -># rounded to three sig figs

To test the result, calculate the density of the sugar solution

${\rho}_{\text{sol" = "5.2752 g"/"5.00 mL" = "1.055 g mL}}^{- 1}$

This density is characteristic of a sugar solution that is approximately $\text{14% w/w}$

http://homepages.gac.edu/~cellab/chpts/chpt3/table3-2.html