A 5 mole sample of liquid acetone is converted to a gas at 75.0°C. If 628 J are required to raise the temperature of the liquid
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Answer:
0.172 M
Explanation:
The reaction for the first titration is:
- HCl + NaOH → NaCl + H₂O
First we <u>calculate how many HCl moles reacted</u>, using the <em>given concentration and volume</em>:
- 19.6 mL * 0.189 M = 3.704 mmol HCl
As one HCl mol reacts with one NaOH mol, <em>there are 3.704 NaOH mmoles in 25.0 mL of solution</em>. With that in mind we <u>determine the NaOH solution concentration</u>:
- 3.704 mmol / 25.0 mL = 0.148 M
As for the second titration:
- H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O
We <u>determine how many NaOH moles reacted</u>:
- 34.9 mL * 0.148 M = 5.165 mmol NaOH
Then we <u>convert NaOH moles into H₃PO₄ moles</u>, using the <em>stoichiometric coefficients</em>:
- 5.165 mmol NaOH *
= 1.722 mmol H₃PO₄
Finally we <u>determine the H₃PO₄ solution concentration</u>:
- 1.722 mmol / 10.0 mL = 0.172 M