# If 0,5kg of dextrose is dissolved in 600ml of water to produce a solution with a final volumeof 1000ml what is the percentage w/w strength of the solution?

Feb 15, 2018

=45.5%

#### Explanation:

Since the problem calls for the concentration of the solution in %(ww); the suitable formula to solve it is the following:

color(red)(%(w/w)=("mass solute")/("mass solution")

$w h e r e :$

${\text{mass solute=mass of dextrose}}_{s o l u t e} = 0.5 k g = 500 g$

$\text{mass solution"=} m a s s . s o l u t {e}_{\mathrm{de} x t r o s e} + m a s s . s o l v e n {t}_{w a t e r}$
${\text{mass solution"=0.5kg+"mass of solvent}}_{w a t e r}$

But, given the volume of water and on the assumption that the density of water @${4}^{o} C = 1 \text{g/ml"=1"kg/L}$, the mass of the solvent (water) can be obtained as shown below.

"mass "=("density")("volume")
${\text{mass solvent}}_{w a t e r} = \left(\frac{1 k g}{\cancel{L}}\right) \left(600 \cancel{m L} \times \frac{1 \cancel{L}}{1000 \cancel{m L}}\right)$
$\text{mass} = 0.6 k g$

This time, masses composing the solution are already known; thus,

$\text{mass solution=mass solute+mass solvent}$

$\text{mass solution=mass dextrose+mass water}$

$\text{mass solution} = 0.5 k g + 0.6 k g$

$\text{mass solution} = 1.1 k g$

Therefore:

$1.1 k g \text{ final mass solution"-=1L " final volume solution}$
or
$1100 g \text{ final mass solution"-=1000mL " final volume solution}$
$\rho \text{ solution} = \frac{1.1 g}{m L} = \frac{1.1 k g}{L}$

Using the formula highlighted above, the concentration of the solution is:

color(red)(%(w/w)=("mass solute")/("mass solution")xx100
color(red)(%(w/w)=(0.5cancel(kg))/(1.1cancel(kg))xx100
color(red)(%(w/w)=0.4545xx100
color(red)(%(w/w)=45.5%

Therefore; the concentration of the solution is 45.5%