Substance X is in a chamber (Chamber 1) containing 10L of fluid. The concentration of substance X in this chamber is 100mmol/L. A solid division separates this chamber from another chamber (Chamber 2) containing 5L of fluid. You remove this division leaving only a membrane separating the two chambers. This membrane is permeable to Substance X and nothing else. Which of the following statements are true?There are several answers to this question, but only one is giving me issues. One of the correct answers is: "The concentration of Substance X in Chamber 2 will never exceed 66mmol/L." I do not understand how they got this as I did a ratio and came up with 50 mmol/L instead. Could someone please point out how it is 66?

Explanation:

concentration of substance X in chamber 1 = 100 mmol/L

total volume of chamber 1 = 10 L

total mass of substance X in chamber 1 = 100 x 10 mmol = 1000 mmol .

When the two chamber is joined , total volume of both the chamber

= 10 L + 5 L = 15 L .

In the volume of 15 litre , substance x is uniformly distributed because it is permeable .

concentration of substance X = mass of X / total volume = 1000 mmol / 15

= 66.67 mmol / L

Hence ,

"The concentration of Substance X in Chamber 2 will never exceed 66mmol/L."

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂