Question

Use the data below to calculate the molar mass of an unknown monoprotic acid and submit a formal lab report via the assignment tab. A. Standardization Of NaOH Solution Please show all calculations: Trial 1 0.3555 g Trial 2 0.3545 g 17.46 mL 17.20 ml Mass of KHP Moles of KHP Volume of NaOH added Moles of NaOH Molarity of NaOH Average Molarity of NaOH % Difference between two readings

Answer 1

Trial 1 of NaOH KH (GH KHP + NaOH - Nakp + H₂O Ind of kHP reacts with Ind of NaOH number of males of KMP = number of males of NaOH number of males of kHP = mass = 0.3555g = 0.00174 I molar mars 204.22 g/mol mol Males of NaOH = Volume & Molarity 1- males = 0.01746 L XM 0.00174 mol = 0.01746 L XM - M = 0.00174 nal/0.01746L - 0.0997 M Nach moles of kHP = 0.3545g = 0.001736 mol - 204 .22g/mol = males of NaOH moles of NaOH = gepea 0.01720LXM 0.001736 mol = 0.01720 LXM M = 0.001736 mol - 0.01720 L -0.1009 M Trual? Percent dif illerence e-le, - E2 x 100% = 10.0997-0-100gx 1001 & tez 0.0997 +0.1009 2 2 = 1.2%