Data: Mass of Candy Milk()14229 Concentration of NaOH (M Initial buret reading (
ID: 588904 • Letter: D
Question
Data: Mass of Candy Milk()14229 Concentration of NaOH (M Initial buret reading (ml.) Candy Trial1 Candy Trial 2 Milk Trial 1 Milk Trial 2 Zhe Otf 14299 25 579 4 Final buret reading (ml.) 225L A5uL Volume of NaOH used (mul Iisuel -2nul 415uLL4HowL ed CIN Litas) Calculations: 1. Determine the moles of NaOH used for each substance for each trial. a Candy Triali 0,100M 2,15ML -2.1510 oes , 0027L ,00115L 0040L b. Milk: Triali 0.100x 4.15ML. 415 10 oles Trial 2 0. 100 x 4.40uLs4.4o»D''holes uoles, acid per gru saipe-Hofucks of acid l sapereight ingau ermine the moles of the acid in the candy and milk for each trial. [Hint: At the end point, the moles of the acid equal the moles of the base]. a. Candy: "CCO215 Los l4229; 1,512 x10 Trial 1 3. Determine the moles of acid per gram of candy and milk for each trial. [Hint: Calculate by dividing the moles of acid by the mass of the candy/milk) a. Candy: Trial 1 Trial 2Explanation / Answer
Ans 2) Your second answer is wrong. Actually, the moles of NaOH used is equal to moles of acid used. The moles you calculated for NaOH in answer 1 is same for moles of acid in candy and in milk.
And the calculation you did in answer 2 is actually for answer 3.
Ans 3) Moles of acid per gram of candy;
Trail 1: moles of acid/ gram = 2.15 x 10-4/1.422 = 1.51 x 10-4 mol/g
Trail 2: moles of acid/ gram = 2.7 x 10-4/ 1.429 = 1.89 x 10-4 mol/g
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