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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.
<h3>The density of H₂ = 0.033 g/L</h3><h3>Further explanation</h3>
In general, the gas equation can be written
where
P = pressure, atm , N/m²
V = volume, liter
n = number of moles
R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m², v= m³)
T = temperature, Kelvin
n = N / No
n = mole
No = Avogadro number (6.02.10²³)
n = m / MW
m = mass
MW = molecular weight
For density , can be formulated :
P = 327 mmHg = 0,430263 atm
R = 0.082 L.atm / mol K
T = 48 ºC = 321.15 K
MW of H₂ = 2.015 g/mol
The density :