Mass of Candy/Milk(g)- T. 400a1-13), Concentration of NaOH (M) 00O sad hooa Init

Question

Mass of Candy/Milk(g)- T. 400a1-13), Concentration of NaOH (M) 00O sad hooa Initial buret reading (mL) | Final buret reading (mL) Volume of NaOH used (mL) Tandy Trial 1 Candy Trial21 Milk Trial 1.1-MilkTrial2 .ao Calculations: 1. Determine the moles of NaOH used for each substance for each trial. a. Candy: Trial 1 Trial 2 b. Milk: Trial 1 Trial 2 2. Determine 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: Trial 1 Trial 2 b. Milk: Trial 1 Trial 2 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 3. Trial 1 Trial 2

Explanation / Answer

From the data,

Calculations

1. moles of NaOH = molarity x volume

Candy Trial #1 = 0.09672 M x 0.0032 ml = 3.1 x 10^-4 mol

Candy Trial #2 = 0.09672 M x 0.00195 ml = 2 x 10^-4 mol

Milk Trial #1 = 0.09672 M x 0.00513 ml = 5 x 10^-4 mol

Milk Trial #1 = 0.09672 M x 0.00333 ml = 3.22 x 10^-5 mol

2. moles of acid in,

moles of acid = moles of base

Candy Trial #1 = 3.1 x 10^-4 mol

Candy Trial #2 = 2 x 10^-4 mol

Milk Trial #1 = 5 x 10^-4 mol

Milk Trial #1 = 3.22 x 10^-5 mol

3. moles of acid/g of candy or milk

Candy Trial #1 = 3.1 x 10^-4 mol/1.4 g = 2.2 x 10^-4 mol/g

Candy Trial #2 = 2 x 10^-4 mol/1.34 g = 1.5 x 10^-4 mol/g

Milk Trial #1 = 5 x 10^-4 mol/26.54 g = 2 x 10^-5 mol/g

Milk Trial #1 = 3.22 x 10^-5 mol/18.01 g = 2 x 10^-6 mol/g

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