Question

CHML 201 Post lab 20 Answer the Q.'s on a separate sheet of paper and attach to your lab report. Write your name on the sheet. 1. A 0.2736 g sample of sulfamic acid (H2NSO3H) required 15.55 mL of a NaOH solution for titration to the phenolphthalein end point. What is the molarity of the NaOH solution? H2NSO3H(aq) + NaOH(aq) = H2NSO3Na(aq) + H2O(l) 2. A 0.1136 g sample of sodium carbonate required 10.55 mL of a H2SO4 solution for titration to the methyl orange end point. What is the molarity of the H2SO4 solution? Na2CO3(ag) + H2SO4(ag) = Na2SO4 (ag) +CO2() H2O(1) 3. A 0.4056 g of oxalic acid dihydrate (H2C204.2H20) required 18.35 mL of a potassium hydroxide for neutralization to the phenolphthalein end point. What is the molarity of the KOH solution? H2C204*2H20 (aq) + 2KOH(aq) = K2C2O4 (aq) +4H20(1)

Answer 1

मातारक्त कुछ H2NSO3 H no. of moles o Volume = Normality & I X 3+14 + 32 +16*3 ng IS N x 15.55 ml xL = no. of noles 1000 men Mw=97 gad 0.2736g sulfamic acid of moles in ne = 0.27364. = 2.821) 97 g/mol 759, = 2.821x10-modes Njx15.55-0.821x10 n = n2 Lobo IN = 0.18M| ③ 0.1136 g Na₂CO MW-106 For methyl orange end point, gland = 2 For H₂SO4 , na = 2 IN= M x Na Naa - NV = no. of moles Max2 x 10.55 L = 0.11369 x 2 TODO 106g ud M = 01136 x 1000 x2 106 10.55x2 Ma = 0.05 M x2 Mar O.IM

H, C, O4 + 2H,б Ми з 6x1 + 6 x 16 + 2x 12 = 126 g/-a. o 4056 4 18 - 3S x Hxj 126 4 7-4 1ою м, = 0-4056 x јбоо 126 18:35 |М – 0-175 м