# 68th Zinc Warm Up Question

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Warm up questions: Exercise1： mass of ZnI2 is 2.56g The molar mass of ZnI2 is 319.18 g/mol The mole of ZnI2: 2.56g/319.18g/mol=0.00802mol 0.00802mol/(500*10-3L)=0.01604M 0.01604M ZnI2 should appear on the label of the flask. Exercise 2: Student 1: 0.43g zinc iodide The mole of ZnI2: 0.43g/319.18 g/mol= 0.00135mol 0.00135mol/0.01604M=0.084L 0.084L*(1000mL/1L)=84mL Student 2: 5.0*10-4 moles of zinc iodide. 5.0*10-4mole/0.01604M=0.031L 0.031L*(1000mL/1L)=31mL Molarity as a Concentration Unit Exercise 3: a. 2.56g ZnI2/500 mL of solution 2.56g/319.18 g/mol=0.00802mol 0.00802mol/(500*10-3L)=0.016M b. 0.00512g ZnI2/mL of solution 0.00512g/319.18 g/mol=1.6*10-5 mol 1.6*10-5 mol/(1*10-3L)=0.016M c. 0.00806 moles of ZnI2/500 mL of solution 0.00806mol/(500*10-3)L=0.016M d. 0.0161 moles of ZnI2/L of solution 0.0161mol/1L=0.016M Exercise 4: a. The moles of ZnI2: 0.25M*(250*10-3)L=0.0625mol b. 0.25M*(250*10-3)L=0.0625mol The mass of ZnI2: 0.0625mol*319.18 g/mol=19.95g c. 0.25M*(500*10-3)L=0.125mol 0.125mol*319.18 g/mol=39.9g ZnI2 d. 0.0125mol/0.25M=0.05L Exercise 5: a. 0.125M*(100*10-3)L=0.0125mol b. 0.0625mol/0.125M=0.5L=500mL Calculation for preparing the EDTA solution Exercise 6 a. 1L*0.02M=0.02mol 0.02mol*372.24g/mol=7.4g EDTA b. Exact molarity of 7.4448g /1.00L would be .0200 M Exercise 7 a. 0.5M*100*10-3L=0.05mol acetic acid b. 0.05mol/6M=8.3*10-3 L=8.3mL stock solution c. 100mL-8.3mL=91.7mLwater Add 91.7 water to 6M stock solution to prepare 0.5M acetic acid. Exercise 8: a. 42.35 - 0.55 = 41.8 mL b. The moles of EDTA4- : 0.0189M*(41.8*10-3)L=7.9*10-4mol c. Zn2+(aq)+EDTA4-(aq)—Zn(EDTA)2-(aq) The ratio Zn2+ and EDTA4- is 1:1 The moles of Zn2+= the moles of EDTA4-=7.9*10-4mol d. 7.9*10-4mol*65.39g/mol=0.0517g Zn e.