# Question 278e0

Feb 18, 2015

You would need to add $794$ $\mu L$ of the $\text{32%}$ solution.

So, you know that your final solution is $\text{40%}$ 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

${V}_{\text{ethanol}} = \frac{40 \cdot 900 \mu L}{100} = 360$ $\mu L$

Now, this amount of ethanol must come from two sources: the absolute ethanol - I'll label this ${V}_{\text{100}}$ - and from the $\text{32%}$ solution - I"ll label this ${V}_{\text{solution 32}}$. You can therefore set up two equations

${V}_{\text{ethanol") = V_("100") + V_("ethanol 32}}$ (1), and

${V}_{\text{solution") = V_("100") + V_("solution 32}}$ (2)

Now, you can express the volume of $\text{32%}$ 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

${V}_{\text{solution") = V_("100") + V_("ethanol 32}} \cdot \frac{100}{32}$

So,

$360 = {V}_{\text{100") + V_("ethanol 32}}$
$900 = {V}_{\text{100") + V_("ethanol 32}} \cdot \frac{100}{32}$

You have ${V}_{\text{100") = 360 - V_("ethanol 32}}$, and

$900 = 360 - {V}_{\text{ethanol 32") + V_("ethanol 32}} \cdot \frac{100}{32}$

Solving this will get you ${V}_{\text{ethanol 32}} = \frac{540}{2.215} = 254.1$ $\mu L$

This value represents the amount of ethanol you must get from the $\text{32%}$ solution, which implies that the total volume for that solution must be

${V}_{\text{solution 32") = V_("ethanol 32}} \cdot \frac{100}{32} = 254.1 \cdot \frac{100}{32} = 794$ $\mu L$

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

${V}_{\text{solution 32}} = 800$ $\mu L$

Feb 18, 2015

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

$\text{900 µL" × 40 % = x" µL × 100 %" + y" µL × 32 %}$

Cancelling units, we get

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

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#

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

$y = 790$

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.