Question

CHE 101 EXPERIMENT 8 Page 131 B REPORT FORM: CONCENTRATION UNITS AND PREPARATION OF SOLUTIONS Name Lab Day & Time Date - Partner's Name (10 Points) Part 1: Calculations Involving Concentration Units: Anyone completing more than formulas before lab will be given blank pages to work with. Show ALL calculations. A mixture is prepared from 55.0 g of hexane (C6H14) and 95.0 g of heptane (C7H16). Calculate: a. The weight percent of each component. Total mass of solution b. The mole fraction of each component (include mole calculations and total number of moles)

Answer 1

wight .f.opmass of component totul mass s DATE a. Moss of Hexone = 55 grom mass of Heptane = 95 gram total mass of Solution = 55+95= 15 ogrom. weight / Of Hexane = = 36.64 Weight % of Heptance - 95 zx100 - 63.33% 150g = 63.33%. 554 = 0.366X100 Sogram mast 95g molormass too gloof 0.95 Mole of Heptane = e mole of Hexane = dep mass – 554 molar mass 86 g/mol 0.64 CON CIRC Heptance C7H16 = 12x7+16 = 84+1 = 100 Kexome CH14 = 12X6 +14 = 72 +14 = 86 totul mole Number = 0.64 +0.95 = 1.59 Mole fraction mol of Hex and 0.64 To for Hexame =- 0.95 to 64 total mol 0.64 1.59 X Hexone = 0.40

Solvent = Hebtame = 95g solute & nexane - 5 groot mole fraction for Hepteme. - - 0.95 0.95 =10.60 X Hepterne = 1.59 Molatit ay mole of soluto mass of solvent inkg m = 559/86 g/mol m = 95 /1000 kg m = 0.6 4 mol 0.095 kg - 6.74 mol/kg Ans d. Normality - Molarity XR n = Basicity of H₂ po4 To molarity of H3PO4 = 1.225 Three Replaceable Hydrogen so n=3 Normality = 1.225 X 3 Ans 3.675 N OH

DATE DATE DITT Molority Equation M = 1.225M Miv, = M₂ V V , = ? M₂ - 0.02500M 1.225 XV, = 0.02560 x il 12:11 V = 0.02500 X] 0.0204L 1.225 Volume in me = 0.0204 x 1000 = 20.40 ml we need 20.40 me (around) Thanks hope this will help you. Good luck