Question

How would kne slove for part C wjth the given data?

DATA A. NaOH Preparation Volume of concentrated NaOH Molarity of concentrated NaOH 16 mL 6.0M 185 mL Volume of DI water B. NaOH Standardization Trial # 1 1.724, 2 3 1.956 9 1.8799 Mass of KHP Titration with NaOH: Initial buret reading 0.4 I mL 18.99mL 0.400 0.51 mL 20.98 m2 20.22 mb Final buret reading C. Molar Mass of NH X Salt UNKNOWN NUMBER: Mass of unknown 0-033 O OBA6g 0.3999 0.025 L 0.025 0.4309 0.0252 Volume of NaOH Titration with H.CO Initial buret reading 0.4mL 15.71 mL 0.00 0.41 mL ostat 15.51 mL AL 0.2500 M 0.31 mL 14.99 mL Final buret reading Molarity of H.CO. 103

Answer 1

Part C Reaction: NH,X+ NaOH NaX + NH,OH The expression is used to calculate the number of moles: moles = M x V Trial #1 V of NaOH = 0.0250 L M of NaOH = 0.4662 M moles of NaOH added = 0.4662 mol/E x 0.025 E = 0.011655 moles V of H,C,0g = 0.0153 L M of H,C,0% = 0.250 M moles 20 cer 0.250 mol/EX 0.0153 E = 0.003825 moles mol of NaOH reacted with NH_X = moles of NaOH added = 0.011655 moles + NaX + NH2OH NH3 + NaOH moles NH X = moles of NaOH 0.011655 moles NH X = moles of NaOH mol of NH X in sample = 0.01165 mol Mass of sample : 0.396 g For the calculation of the molar mass the expression is used: mass NH X molar mass NH X = moles NH, X molar mass NH X = 0.396 g 0.01165 mol molar mass NH X = 34.00

Trial # 2 V of NaOH = 0.0250 L M of NaOH = 0.4654 M moles of NaOH added = 0.4654 mol/E x 0.025 £ = 0.011635 moles V of H2C2O3 = 0.0151 L Mof H,C,08 = 0.250 M moles of H,C,Og used= 0.250 mol/L x 0.0151 E = 0.003775 moles mol of NaOH reacted with NH_X = moles of NaOH added = 0.011635 moles NaX + NH,OH NH,X+ NaOH - moles NH X = moles of NaOH 0.011635 moles NH X = moles of NaOH mol of NH X in sample = 0.011635 mol Mass of sample : 0.399 g of the molar mass the For the calculation expression is used: molar mass NH X = mass NH X moles NH X molar mass NH X = 0.399 g 0.011635 mol molar mass NH X = 34.29 g/mol

Trial # 2 V of NaOH = 0.0250 L M of NaOH = 0.4654 M moles of NaOH added = 0.4654 mol/E x 0.025 £ = 0.011635 moles V of H2C2O3 = 0.0151 L Mof H,C,08 = 0.250 M moles of H,C,Og used= 0.250 mol/L x 0.0151 E = 0.003775 moles mol of NaOH reacted with NH_X = moles of NaOH added = 0.011635 moles NaX + NH,OH NH,X+ NaOH - moles NH X = moles of NaOH 0.011635 moles NH X = moles of NaOH mol of NH X in sample = 0.011635 mol Mass of sample : 0.399 g of the molar mass the For the calculation expression is used: molar mass NH X = mass NH X moles NH X 0.399 g molar mass NH X = 0.011635 mol molar mass NHX = 34.29 g/mol 34.0 g/mol + 34.0 /mol + 37.0 /mol Average NH X molar mass = Average NH X molar mass = 35.1 /mol )