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?

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

Explanation:

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

× = ×

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

= is our unknown value but we know the other three values in the equation:

= 95%

= 285 mL

= 70%

So, clearing the previous given formula we obtain:

=

and replacing for numerical values:

= 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.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%