14. 60. g of naoh is dissolved in enough distilled water to make 300 ml of a stock solution. what volumes of this solution and distilled water, when mixed, will result in a solution that is approximately 1 m naoh?

The question is incomplete, the complete question is attached below.

Answer : The volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.

Explanation : Given,

Mass of NaOH = 60 g

Volume of stock solution = 300 mL

Molar mass of NaOH = 40 g/mol

First we have to calculate the molarity of stock solution.

Now put all the given values in this formula, we get:

Now we have to determine the volume of stock solution and distilled water mixed.

Formula used :

where,

are the molarity and volume of stock solution.

are the molarity and volume of diluted solution.

From data (A) :

Putting values in above equation, we get:

Volume of stock solution = 20 mL

Volume of distilled water = 100 mL - 20 mL = 80 mL

From data (B) :

Putting values in above equation, we get:

Volume of stock solution = 20 mL

Volume of distilled water = 100 mL - 20 mL = 80 mL

From data (C) :

Putting values in above equation, we get:

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From data (D) :

Putting values in above equation, we get:

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From this we conclude that, when 20 mL stock solution and 80 mL distilled water mixed then it will result in a solution that is approximately 1 M NaOH.

Hence, the volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.

it's filtration, but this is an assumption so my apologies if i got this