Question

Question #ca68c

Answer 1

You'd have to use 5 mL of 5 M sucrose and 95 mL of water.

A very useful tool to have when doind dilution calculations is the dilution factor - read more here:

http://socratic.org/questions/how-do-you-calculate-dilution-factor?source=search

The dilution factor will tell you by what factor you've diluted the original sample. In your case, you need to go from a concentration of 5 M to a concentration of 0.25 M.

You need the dilution factor to be equal to 20, since a molarity of 0.25 M is 20 times smaller than a molarity of 5 M.

The dilution factor is defined as the final volume of the solution divided by the initial volume of the sample

`"DF" = V_"finaL"/V_"initial"`

Check to see which of your solutions would give you a dilution factor of 20

`"DF"_1 = ((0.05 + 99.95)cancel("mL"))/(0.05cancel("mL")) = 2,000`

This volume ratio is too high.

`"DF"_2 = ((0.5 + 99.5)cancel("mL"))/(0.5cancel("mL")) = "200"`

The volume ratio is till too high, move on to the next solution.

`"DF"_3 = ((5 + 95)cancel("mL"))/(5cancel("mL")) = color(green)(20)`

There it is. If you mix 5 mL of a 5-M sucrose solution with enough water to make the final volume equal to 100 mL (with 95 mL of water, to be precise), you'd get 100 mL of a 0.25-M solution. solution