Suppose that you have a 60.0% solution of NaOH. How many milliliters of water must be added to 30.0 mL of this solution to prepare a 35.0% solution of NaOH

21.4 mL

Explanation:

Firstly, one has to understand some "terms" here before proceeding.

60.0% solution of NaOH means 60g of NaOH salt/solute in 100 mL of water

35.0% solution of NaOH also means 35g of NaOH salt/solute in 100 mL of water.

To solve the problem in the question, we say, if 60 g dissolves in 100 mL, how many grams will be present in 30 mL

60 ⇒ 100

? ⇒ 30

? = 60 × 30/100

? = 18g

18 g of NaOH will be present in 30 mL of the solution making the % weight per volume to be 18/30

Thus, if we want to achieve 35% solution of NaOH from 18/30, the total amount of water that will be required will be ?

35 ⇒ 100

18 ⇒ ?

? = 18 × 100/35

? = 51.4 mL

Hence, the milliliters of water that must be added is 51.4 - 30 = 21.4 mL

your answer would be “dry”

answer; the numbers of valence electrons in the acetic acid molecule (ch3co2h) is 24;