A 15.0 ml sample of gas at 10.0 degree Celsius and 760 torr changes to a pressure of 1252 torr at 35.0 degree Celsius. What is t
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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.
