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.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3