A 0.573 g portion of an ammonium salt containing a single mole of ammonium ion for each mole of the salt was reacted with 50.00 mL of 0.2000 M NaOH. After the reaction the solution was boiled to remove ammonia created during the reaction. The resulting solution was back titrated to the end point with 26.80 mL of 0.1000 M H_2C_2O_4. 1. Calculate the number of moles of hydroxide ion in the 50.00 mL of 0.2000 M sodium hydroxide solution added to the ammonium salt. 2. From the volume and concentration of oxalic acid used during the back titration, calculate the number of moles of oxalic acid used. 3. By using the balanced chemical equation for the back titration [Reaction (14-2)1 and the number of moles of oxalic acid from step 2, calculate the number of moles of excess hydroxide. 4. Using the number of moles of sodium hydroxide from step 1 (the moles of hydroxide initially added to the sample) and the number of moles of excess hydroxide from step 3, calculate the number of moles of hydroxide that reacted with the ammonium salt. 5. Using the number of moles of sodium hydroxide that reacted with the ammonium salt (the result of the calculation in step 4), and the balanced chemical equation for the reaction between hydroxide and the ammonium [Reaction (14-1)], what is the number of moles of ammonium ion in the 0.573 g sample? 6. Since each mole of salt contains one mole of ammonium ion, the answer obtained in step 5 is also the number of moles of ammonium salt. Use the mass of the sample (g) and the number of moles of salt in the sample to calculate the molar mass of the ammonium salt (g/mol).