Question

3. You are preparing 285 mL of 70% (v/v) ethanol in water using a 95% ethanol stock. Round your answer to the nearest whole number. (Ch. 2.4) a) How much 95% ethanol (in mL) do you use? b) How much water (in mL) do you add?

Answer 1

The answer is for a) 210 mL of ethanol b) 75 mL of water.

Step-by-step explanation:

When we prepare a solution using a stock solution we use the following equation:

`V_(1)`×
`C_(1)` =
`V_(2)`×
`C_(2)`

This is an equation for dilutions and we can use it in the case of our answer.

`V_(1)` = is our unknown value but we know the other three values in the equation:

`C_(1)` = 95%

`V_(2)` = 285 mL

`C_(2)` = 70%

So, clearing the previous given formula we obtain:

`(C_(2) V_(2) )/(C_(1) )` =
`V_(1)`

and replacing for numerical values:

`(70x285)/(95)` = 210 mL this is the value of ethanol stock solution we need to use

and subtracting this value of ethanol from the total volume of the solution

285 mL - 210mL = 75 mL of water needed.

Answer 2

a) 210 mL of 95% ethanol

b) 75 mL of water

Step-by-step explanation:

a) We can use the dilution equation to solve this problem. C₁ and V₁ are the concentration and volume of the stock solution, respectively, while C₂ and V₂ are the concentration and volume of the diluted solution.

C₁V₁ = C₂V₂

We want to find V₁, the volume of ethanol stock we need to dilute to prepare the final solution:

V₁ = (C₂V₂) / C₁

The concentrations are represented in percentages. We substitute in the known values to calculate V₁. The units cancel to leave us with units of mL.

V₁ = (C₂V₂) / C₁ = (70%)(285mL) / (95%) = 210 mL

b) The final solution volume is 285 mL and we have added 210 mL of ethanol, so the remaining volume is from the water that we add:

(285 mL - 210 mL) = 75 mL