Given:
M = 0.0150 mol/L HF solution
T = 26°C = 299.15 K
π = 0.449 atm
Required:
percent ionization
Solution:
First, we get the van't Hoff factor using this equation:
π = i MRT
0.449 atm = i (0.0150 mol/L) (0.08206 L atm / mol K) (299.15 K)
i = 1.219367
Next, calculate the concentration of the ions and the acid.
We let x = [H+] = [F-]
[HF] = 0.0150 - x
Adding all the concentration and equating to iM
x +x + 0.0150 - x = <span>1.219367 (0.0150)
x = 3.2905 x 10^-3
percent dissociation = (x/M) (100) = (3.2905 x 10-3/0.0150) (100) = 21.94%
Also,
percent dissociation = (i -1) (100) = (</span><span>1.219367 * 1) (100) = 21.94%</span>
Answer:
3.67 moles of N
Explanation:
The epinephrine's chemical formula is: C₉H₁₃O₃N
We were told that a chemist found that in a mesaure of epinephrine, he found 33 moles of C
We must know that 9 moles of C are in 1 mol of C₉H₁₃O₃N so, let's make a rule of three:
If 9 moles of C are found in 1 mol of C₉H₁₃O₃N
Therefore 33 moles of C must be found in (33 .1) / 9 = 3.67 moles of C₉H₁₃O₃N
There is a second rule of three, then.
In 1 mol of C₉H₁₃O₃N we have 1 mol of N
Then, 3.67 moles C₉H₁₃O₃N must have (3.67 . 1) / 1 = 3.67 moles of N
Remember 1 mol of C₉H₁₃O₃N has 9 moles of C, 13 moles of H, 3 moles of O and 1 mol of N
Take a hypothetical sample of exactly 100 grams of the solution.
(16g urea) / (60.06 g urea/mol) = 0.2664 mol urea
((100 g total) - (16g urea)) = 84.0 g H2O = 0.0840 kg H2O
(0.2664 mol) /0.0840 (kg) = 3.17143mol/kg = 3.18m urea
