Question #278e0

You would need to add #794# #muL# of the #"32%"# solution.

So, you know that your final solution is #"40%"# ethanol, which means you can use its volum to determine exactly how much ethanol it should contain

#"%40" = V_("ethanol")/V_("solution") * 100 => V_("ethanol") = (40 * V_("solution"))/100#

#V_("ethanol") = (40 * 900 muL)/100 = 360# #muL#

Now, this amount of ethanol must come from two sources: the absolute ethanol - I'll label this #V_("100")# - and from the #"32%"# solution - I"ll label this #V_("solution 32")#. You can therefore set up two equations

#V_("ethanol") = V_("100") + V_("ethanol 32")# (1), and

#V_("solution") = V_("100") + V_("solution 32")# (2)

Now, you can express the volume of #"32%"# solution like this

#"%32" = V_("ethanol 32")/V_("solution 32") * 100 => V_("solution 32") = V_("ethanol 32") * 100/32#

Plug this into equation (2) and you'll get

#V_("solution") = V_("100") + V_("ethanol 32") * 100/32#

So,

#360 = V_("100") + V_("ethanol 32")#
#900 = V_("100") + V_("ethanol 32") * 100/32#

You have #V_("100") = 360 - V_("ethanol 32")#, and

#900 = 360 - V_("ethanol 32") + V_("ethanol 32") * 100/32#

Solving this will get you #V_("ethanol 32") = 540/2.215 = 254.1# #muL#

This value represents the amount of ethanol you must get from the #"32%"# solution, which implies that the total volume for that solution must be

#V_("solution 32") = V_("ethanol 32") * 100/32 = 254.1 * 100/32 = 794# #muL#

If you round this value to one sig fig, the answer will be

#V_("solution 32") = 800# #muL#

The student should use 790 µL of 32 % ethanol.

Let #x# = the volume of 100 % ethanol and #y# = the volume of 32 % ethanol.

Then we can re-write the problem as

#"900 µL" × 40 % = x" µL × 100 %" + y" µL × 32 %"#

Cancelling units, we get

(1) #36 000 = 100x + 32y#

We also know that the total volume must be 900 µL.

(2) #x + y = 900#

We have two equations in two unknowns.

Solve (2) for #x#.

(3) #x = 900 – y#

Substitute (3) into (1).

#36 000 = 100(900 – y) + 32y= 90 000 – 100 y + 32 y = 90 000 - 68y#

#68y = 90 000 -36 000 = 54 000#

#y = 790#

So you would use 790 µL of 32 % ethanol.

Note: The answer can have only 2 significant figures, because that is all you gave for the
32 % ethanol. If you need more precision, you will have to recalculate.