Answer: Option (b) is the correct answer.
Explanation:
The given data is as follows.
mass = 0.508 g, Volume = 0.175 L
Temperature = (25 + 273) K = 298 K, P = 1 atm
As per the ideal gas law, PV = nRT.
where, n = no. of moles = 
Hence, putting all the given values into the ideal gas equation as follows.
PV = 
1 atm \times 0.175 L = 
= 71.02 g
As the molar mass of a chlorine atom is 35.4 g/mol and it exists as a gas. So, molar mass of 
is 70.8 g/mol or 71 g/mol (approx).
Thus, we can conclude that the gas is most likely chlorine.
The answer is 35.4335
Hope this helped! (Plz mark me brainliest!)
Answer:
2.5L [NaCl] concentrate needs to be 4.8 Molar solution before dilution to prep 10L of 1.2M KNO₃ solution.
Explanation:
Generally, moles of solute in solution before dilution must equal moles of solute after dilution.
By definition Molarity = moles solute/volume of solution in Liters
=> moles solute = Molarity x Volume (L)
Apply moles before dilution = moles after dilution ...
=> (Molarity X Volume)before dilution = (Molarity X Volume)after dilution
=> (M)(2.5L)before = (1.2M)(10.0L)after
=> Molarity of 2.5L concentrate = (1.2M)(10.0L)/(2.5L) = 4.8 Molar concentrate
