Question #f3cad

1 Answer
Mar 15, 2017

Answer:

#0.94%#

Explanation:

I will assume that the glucose solution is being added to #"375 mL"# of water.

The idea here is that the concentration of the solution will decrease because the mass of solute remains constant while the volume of solvent increases.

Your initial solution had a concentration of #"15% m/v"#, which basically means that you get #"15 g"# of glucose, the solute, for every #"100 mL"# of solution.

Now, you don't have to calculate the amount of solute present in the initial solution, all you have to do here is figure out what the total volume of the solution will be after the dilution.

#V_"total" = "25 mL" + "375 mL" = "400 mL"#

You can say that the initial solution is being diluted by a factor of

#"DF" = (400 color(red)(cancel(color(black)("mL"))))/(25color(red)(cancel(color(black)("mL")))) = color(blue)(16) -># this is the dilution factor

This means that the diluted solution contains #color(blue)(16)# times more solvent and the same amount of solute as the initial solution contained.

Therefore, you can say that the concentration of the diluted solution is #color(blue)(16)# times lower than that of the initial solution

#"% diluted" = "15%"/color(blue)(16) = color(darkgreen)(ul(color(black)(0.94%)))#

The answer is rounded to two sig figs.