See attachement for the answer to your question.
Answer:
The answer would be at 120°C
Answer:
B) 0.32 %
Explanation:
Given that:

Concentration = 1.8 M
Considering the ICE table for the dissociation of acid as:-

The expression for dissociation constant of acid is:
![K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}](https://tex.z-dn.net/?f=K_%7Ba%7D%3D%5Cfrac%20%7B%5Cleft%20%5B%20H%5E%7B%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20%7BCH_3COO%7D%5E-%20%5Cright%20%5D%7D%7B%5BCH_3COOH%5D%7D)


Solving for x, we get:
<u>x = 0.00568 M</u>
Percentage ionization = 
<u>Option B is correct.</u>
We shall find the molar mass first.
Ca5(PO4)3(OH) = (40 * 5) + 3 (31 + 4(16)) + 16 + 1 = 200 + 285 + 17 = 485 + 17 = 502.
Percent of Calcium = 200/ 502 * 100 = 39.8%
Percent of Phosphorus = 91/502 * 100 = 18.1%
Percent of Hydrogen = 1/502 * 100 = 0.19%
Percent of Oxygen = 100 - (39.8 + 18.1 + 0.19) = 41.91%