The asnwer is #2.04%# for the **w/v** solution and #1.94%# for the **v/v** solution.

Let's start by clearly defining what our solutions look like.

A **w/v** solution is usually used when a dry chemical is mixed as a dry mass (grams) per volume (mL), where

#(n umber.of.grams)/(100mL)# = **percent concentration**

So, a #5%# **w/v** solution will have #5g# of a substance in #100mL# of ethanol.

A **v/v** solution is used when dealing with liquid reagents and it represents

#(n umber.of.mL)/(100mL)# = **percent concentration**

So, a #5%# **v/v** solution will have #5mL# of a substance in #95mL# ethanol.

Now, let's start with the **w/v** solution. An initial #5%# concentration is now brought to #0.1%#; this means that the volume of the solution is now 50 times larger (since the concentration is 50 times smaller).

Therefore, the #5g# of the substance would now occupy

#50 * 100mL = 5000mL #, out of which only the initial #100mL# would represent ethanol (since the initial solutions are diluted in saline). The percentage of ethanol in the final solution would be

#(100mL)/((5000 - 100)mL) * 100% = 2.04%#

The procedure is similar for the **v/v** solution; the initial #5%# solution's volume will now be 50 times larger, which means that the #5g# of substance will now occupy

#50 * 100mL = 5000mL#, out of which #95mL# would represent ethanol. The percentage of ethanol would now be

#(95mL)/((5000 - 95)mL) * 100% = 1.94%#