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.