NAME A sample of 0.231 g of an unknown acid is dissolved in water and is titrate

Question

NAME A sample of 0.231 g of an unknown acid is dissolved in water and is titrated with a 0.117 M NaOH(ag solution. Find the molar 9. A sample of 0. mass of the 6pts m to reach the equivalence point. ch the equivalence point.h uknown monoprotic acid if 26.3 ml of the NaOH(aq) solution is required Answer:- g/mol . A gas fuel is combusted into an expandable cylinder releasing 222 kJ of heat into the environment. During eprocess the piston expands and the gases perform 42 kJ of pressure-volume work. Find (a) the internal rgy change and (b) the enthalpy change of the gases in the piston. 6pts

Explanation / Answer

9)
At equivalence point, the moles of NaOH added is equal to the moles of acid present.

Moles of NaOH = Volume NaOH in L x molarity of NaOH

                          = (26.3 mLx 1 L/1000 mL) x (0.117 M)

                         = 0.0263 L x (0.117 mol/L)

                        = 0.003077 mol

So, the moles of unknwon acid is 0.003077 mol.

Mass of unknown acid is 0.231 g.

Molar mass of unknown acid = mass of acid/moles of acid

                                              = 0.231 g/0.003077 mol

                                              = 75.1 g/mol

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.